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expression to $\\dfrac{2}{1 - \\cos^2 \\phi}$ A1\n- Uses the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$ to obtain $\\dfrac{2}{\\sin^2 \\phi}$ and concludes $2 \\csc^2 \\phi$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"lhs_to_rhs","label":"LHS to RHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we start with the expression on the Left-Hand Side (LHS) of the identity:\n\n$$\\text{LHS} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{LHS} = \\frac{\\htmlData{anchor=num1}{\\color{@sky}{1(1 + \\cos \\phi)}} + \\htmlData{anchor=num2}{\\color{@pink}{1(1 - \\cos \\phi)}}}{\\htmlData{anchor=denom}{\\color{@purple}{(1 - \\cos \\phi)(1 + \\cos \\phi)}}}$$"}],"highlights":[{"color":"sky","style":"underline","anchor":"num1"},{"color":"pink","style":"underline","anchor":"num2"},{"color":"purple","style":"underline","anchor":"denom"}],"connections":[]},"order":1,"title":"Find a common denominator","description":"To combine these two algebraic fractions, find a common denominator by multiplying the denominators: $(1 - \\cos \\phi)(1 + \\cos \\phi)$. This earns the first mark."},{"type":"text","order":2,"title":"Simplify the numerator","content":"Now, we expand and simplify the terms in the numerator:\n\n$$(1 + \\cos \\phi) + (1 - \\cos \\phi) = 1 + 1 + \\cos \\phi - \\cos \\phi = 2$$\n\nSo the fraction becomes:\n\n$$\\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$(1 - \\cos \\phi)(1 + \\cos \\phi) = 1^2 - \\cos^2 \\phi$$"},{"latex":"$$$\\text{LHS} = \\frac{2}{\\htmlData{anchor=diff-sq}{\\color{@teal}{1 - \\cos^2 \\phi}}}$$"}],"highlights":[{"color":"teal","style":"box","anchor":"diff-sq"}],"connections":[]},"order":3,"title":"Expand the denominator","description":"We expand the denominator using the difference of two squares identity, $(a-b)(a+b) = a^2 - b^2$, which simplifies our expression to a single term. This gets us the second mark."},{"type":"text","order":4,"title":"Apply Pythagorean identity","content":"Recall the fundamental Pythagorean trigonometric identity from our syllabus:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1$$\n\nRearranging this gives:\n\n$$1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstituting this into our denominator, we get:\n\n$$\\frac{2}{\\sin^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\csc \\phi = \\frac{1}{\\sin \\phi} \\implies \\htmlData{anchor=recip-id}{\\color{@amber}{\\csc^2 \\phi}} = \\htmlData{anchor=sin-id}{\\color{@amber}{\\frac{1}{\\sin^2 \\phi}}}$$"},{"latex":"$$$\\text{LHS} = 2 \\left( \\color{@amber}{\\frac{1}{\\sin^2 \\phi}} \\right) = 2\\color{@amber}{\\csc^2 \\phi} = \\text{RHS}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sin-id"},{"color":"amber","style":"text-color","anchor":"recip-id"}],"connections":[{"to":"recip-id","from":"sin-id","color":"amber","label":null,"style":"solid"}]},"order":5,"title":"Use reciprocal identity","description":"Since the cosecant function is defined as the reciprocal of sine, we can replace the sine term in the denominator. This complete logical step awards the final mark."}]},{"id":"rhs_to_lhs","label":"RHS to LHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using the RHS to LHS manipulation approach, we begin with the right-hand side of the identity:\n\n$$\\text{RHS} = 2 \\csc^2 \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = 2 \\htmlData{anchor=cosec}{\\csc^2 \\phi}$"},{"text":"We use the fundamental reciprocal identity:"},{"latex":"$$\\htmlData{anchor=recip-def}{\\csc \\phi = \\frac{1}{\\sin \\phi}}$"},{"text":"Substituting this in gives:"},{"latex":"$$\\phantom{\\text{RHS}} = \\htmlData{anchor=sin-sub}{\\frac{2}{\\sin^2 \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cosec"},{"color":"sky","style":"text-color","anchor":"recip-def"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"recip-def","from":"cosec","color":"sky","label":"reciprocal","style":"solid"},{"to":"sin-sub","from":"recip-def","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Convert cosecant to sine","description":"Convert the cosecant term to its equivalent expression in terms of sine."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2}{\\htmlData{anchor=sin-denom}{\\sin^2 \\phi}}$"},{"text":"Using the Pythagorean identity:"},{"latex":"$$\\htmlData{anchor=pythag}{\\sin^2 \\phi + \\cos^2 \\phi = 1} \\implies \\htmlData{anchor=sin-isolated}{\\sin^2 \\phi = 1 - \\cos^2 \\phi}$"},{"text":"We substitute this into our denominator:"},{"latex":"$$\\phantom{\\text{RHS}} = \\frac{2}{\\htmlData{anchor=cos-denom}{1 - \\cos^2 \\phi}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sin-denom"},{"color":"amber","style":"text-color","anchor":"pythag"},{"color":"amber","style":"text-color","anchor":"cos-denom"}],"connections":[{"to":"pythag","from":"sin-denom","color":"amber","label":"identity","style":"solid"},{"to":"cos-denom","from":"sin-isolated","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply Pythagorean identity","description":"Rewrite the denominator using the Pythagorean identity."},{"type":"text","order":3,"title":"Factor the denominator","content":"The expression in the denominator, $1 - \\cos^2 \\phi$, is a difference of two squares. We can factor it as:\n\n$$1 - \\cos^2 \\phi = (1 - \\cos \\phi)(1 + \\cos \\phi)$$\n\nThis updates our fraction to:\n\n$$\\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":4,"title":"Split the fraction","content":"To rewrite this single fraction as two separate fractions, we decompose the numerator. Notice that we can write $2$ as:\n\n$$2 = (1 + \\cos \\phi) + (1 - \\cos \\phi)$$\n\nNow we substitute this into the numerator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nWe can now split this into two separate terms:\n\n$$\\frac{1 + \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)} + \\frac{1 - \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":5,"title":"Simplify to LHS","content":"By cancelling the common factors in the numerator and denominator of each fraction, we get:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = \\text{LHS}$$\n\nSince we have shown that $\\text{RHS} \\equiv \\text{LHS}$, the identity is proven."}]}],"generatedAt":"2026-06-09T03:46:41.518Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"297bfcd9-44d4-443c-b4e0-8c24a73e27a2","label":"b","content":"In a study of robotic arm resonance, the stability factor \$k\$ is defined by the equation\n\\[ \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = k \\]\nDetermine the set of values of \$k\$ for which this equation has real solutions for \$\\phi\$. Fully justify your answer.","markscheme":"- Recognizes the relationship between the identity and the range of sine: $k = 2 \\csc^2 \\phi = \\dfrac{2}{\\sin^2 \\phi}$ M1\n- States that for real solutions, $0 < \\sin^2 \\phi \\le 1$ since $\\cos \\phi \\neq \\pm 1$ M1\n- Concludes that $\\csc^2 \\phi \\ge 1$, therefore the set of values is $k \\ge 2$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"trigonometric_simplification","label":"Trigonometric Simplification","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"To solve for $k$, we start by combining the two fractions on the left-hand side under a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = k$$\n\nMultiply the numerator and denominator of each fraction to get a common denominator of $(1 - \\cos \\phi)(1 + \\cos \\phi)$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = k$$"},{"type":"text","order":1,"title":"Use trigonometric identity","content":"Simplify the numerator and expand the denominator:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = k$$\n\nUsing the fundamental Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$, we can substitute $1 - \\cos^2 \\phi = \\sin^2 \\phi$ in the denominator:\n\n$$k = \\frac{2}{\\sin^2 \\phi}$$\n\nThis earns you the first mark (**M1**) for simplifying the relation."},{"type":"text","order":2,"title":"Determine domain restrictions","content":"For the original expression to be defined, the denominators cannot be zero:\n\n$$1 - \\cos \\phi \\neq 0 \\implies \\cos \\phi \\neq 1$$\n$$1 + \\cos \\phi \\neq 0 \\implies \\cos \\phi \\neq -1$$\n\nSince $\\cos \\phi \\neq \\pm 1$, it follows that $\\sin \\phi \\neq 0$.\n\nKnowing that for any real angle $\\phi$, the range of the sine function is $-1 \\le \\sin \\phi \\le 1$, squaring both sides gives:\n\n$$0 \\le \\sin^2 \\phi \\le 1$$\n\nBecause $\\sin \\phi \\neq 0$, this simplifies to the restricted range:\n\n$$0 < \\sin^2 \\phi \\le 1$$\n\nThis justification earns you the second mark (**M1**)."},{"type":"text","order":3,"title":"Find the range of k","content":"Now we use the inequality $0 < \\sin^2 \\phi \\le 1$ to find the range of $k$:\n\nTaking the reciprocal of both sides reverses the inequality sign:\n\n$$\\frac{1}{\\sin^2 \\phi} \\ge 1$$\n\nMultiplying by $2$:\n\n$$k = \\frac{2}{\\sin^2 \\phi} \\ge 2$$\n\nThus, the set of values of $k$ for which the equation has real solutions is:\n\n$$k \\ge 2$$\n\nThis final conclusion earns you the final accuracy mark (**A1**)."}]},{"id":"algebraic_substitution","label":"Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Introduce algebraic substitution","content":"To solve this problem using algebraic substitution, we define a new variable for the trigonometric term. Let:\n\n$$u = \\cos \\phi$$\n\nFor the fractions in the equation to be defined, the denominators cannot be zero. Therefore, $\\cos \\phi \\neq 1$ and $\\cos \\phi \\neq -1$.\n\nSince the range of the cosine function is $-1 \\le \\cos \\phi \\le 1$, excluding the boundary values gives the domain constraint for $u$:\n\n$$-1 < u < 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{1}{1 - \\htmlData{anchor=u-sub1}{\\color{@sky}{u}}} + \\frac{1}{1 + \\htmlData{anchor=u-sub2}{\\color{@sky}{u}}} = k$$"},{"latex":"$$$\\frac{(1 + \\color{@sky}{u}) + (1 - \\color{@sky}{u})}{(1 - \\color{@sky}{u})(1 + \\color{@sky}{u})} = k$$"},{"latex":"$$$\\frac{2}{1 - \\htmlData{anchor=u-sq}{\\color{@purple}{u^2}}} = k$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-sub1"},{"color":"sky","style":"text-color","anchor":"u-sub2"},{"color":"purple","style":"text-color","anchor":"u-sq"}],"connections":[{"to":"u-sq","from":"u-sub1","color":"purple","label":"combine terms","style":"solid"}]},"order":1,"title":"Simplify the expression","description":"Substitute $u$ into the equation and find a common denominator to simplify the algebraic fractions."},{"type":"text","order":2,"title":"Apply domain constraints","content":"Now we analyze the range of the term $u^2$. Since $-1 < u < 1$, squaring both sides yields:\n\n$$0 \\le u^2 < 1$$\n\nWe can build up the expression for $k$ step-by-step using inequalities:\n\n1. Multiply by $-1$:\n$$-1 < -u^2 \\le 0$$\n\n2. Add $1$ to all parts:\n$$0 < 1 - u^2 \\le 1$$\n\n3. Take the reciprocal (reversing the inequality direction for positive values):\n$$\\frac{1}{1 - u^2} \\ge 1$$\n\n4. Multiply by $2$:\n$$\\frac{2}{1 - u^2} \\ge 2$$"},{"type":"text","order":3,"title":"State the final set","content":"Since $k = \\frac{2}{1 - u^2}$, we conclude that:\n\n$$k \\ge 2$$\n\nThis is the complete set of values for which the equation has real solutions for $\\phi$."}]}],"generatedAt":"2026-06-09T03:47:35.506Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"d766397d-e937-4d6e-a926-38652bba6ec1","label":"c","content":"Given that \$\\phi\$ is in the third quadrant (reflex angle between \$180^\\circ\$ and \$270^\\circ\$) and\n\\[ \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18 \\]\nfind the exact value of \$\\cot \\phi\$.","markscheme":"- Uses the identity from part (a) to set $2 \\csc^2 \\phi = 18$, finding $\\csc^2 \\phi = 9$ B1\n- Employs the identity $1 + \\cot^2 \\phi = \\csc^2 \\phi$ to find $\\cot^2 \\phi = 8$ M1\n- Evaluates $\\cot \\phi = \\pm \\sqrt{8}$ and justifies that $\\cot \\phi = 2\\sqrt{2}$ because $\\cot \\phi$ is positive in the third quadrant A1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"pythagorean_identity","label":"Using Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Simplify the algebraic fraction","content":"First, we can combine the fractions on the left-hand side of the equation by finding a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nMultiply the numerator and denominator of each fraction to combine them:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$\n\nSimplify the numerator and expand the denominator:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 18$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the fundamental identity: } \\htmlData{anchor=trig-identity}{\\color{@sky}{1 - \\cos^2 \\phi}} = \\htmlData{anchor=sin-sq}{\\color{@sky}{\\sin^2 \\phi}}$"},{"latex":"$${\\frac{2}{\\htmlData{anchor=sub-denom}{\\color{@sky}{1 - \\cos^2 \\phi}}}} = 18 \\implies \\frac{2}{\\htmlData{anchor=sin-denom}{\\color{@sky}{\\sin^2 \\phi}}} = 18$"},{"latex":"$$\\text{Since } \\htmlData{anchor=cosec-def}{\\color{@amber}{\\csc^2 \\phi}} = \\htmlData{anchor=inv-sin}{\\color{@amber}{\\frac{1}{\\sin^2 \\phi}}}, \\text{ we can write:}$"},{"latex":"$$2\\htmlData{anchor=cosec-term}{\\color{@amber}{\\csc^2 \\phi}} = 18$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"trig-identity"},{"color":"sky","style":"text-color","anchor":"sin-sq"},{"color":"amber","style":"text-color","anchor":"cosec-def"}],"connections":[{"to":"sub-denom","from":"trig-identity","color":"sky","label":"substitute","style":"solid"},{"to":"cosec-term","from":"cosec-def","color":"amber","label":"rewrite","style":"solid"}]},"order":1,"title":"Express in terms of cosecant","description":"We use the identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$ to rewrite the denominator, and then express the expression in terms of the cosecant function."},{"type":"text","order":2,"title":"Solve for cosecant squared","content":"Now, we divide both sides by $2$ to solve for $\\csc^2 \\phi$:\n\n$$\\csc^2 \\phi = 9$$\n\n*(This earns the first mark B1)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Recall the Pythagorean identity: } \\htmlData{anchor=pythag-id}{\\color{@purple}{1 + \\cot^2 \\phi}} = \\htmlData{anchor=cosec-sq-term}{\\color{@purple}{\\csc^2 \\phi}}$"},{"latex":"$$1 + \\cot^2 \\phi = \\htmlData{anchor=num-val}{\\color{@purple}{9}}$"},{"latex":"$$\\cot^2 \\phi = 8$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"pythag-id"},{"color":"purple","style":"text-color","anchor":"cosec-sq-term"}],"connections":[{"to":"cosec-sq-term","from":"pythag-id","color":"purple","label":"identity","style":"solid"}]},"order":3,"title":"Apply the Pythagorean identity","description":"Using the Pythagorean identity relating cotangent and cosecant, we substitute $\\csc^2 \\phi = 9$ and solve for $\\cot^2 \\phi$."},{"type":"text","order":4,"title":"Quadrant analysis and final value","content":"Taking the square root of both sides gives:\n\n$$\\cot \\phi = \\pm\\sqrt{8} = \\pm 2\\sqrt{2}$$\n\nWe are given that $\\phi$ is in the **third quadrant** ($180^\\circ < \\phi < 270^\\circ$).\n\nIn the third quadrant:\n- Both $\\sin \\phi$ and $\\cos \\phi$ are negative.\n- Therefore, $\\tan \\phi$ and its reciprocal $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$ must be **positive**.\n\nThus, we select the positive root:\n\n$$\\cot \\phi = 2\\sqrt{2}$$\n\n*(Earning M1 for finding $\\cot^2 \\phi = 8$ and A1 for selecting the positive root with correct quadrant justification)*"}]},{"id":"sine_cosine_ratio","label":"Finding Sine and Cosine","steps":[{"type":"text","order":0,"title":"Combine the algebraic fractions","content":"To solve the equation\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nwe first combine the fractions on the left-hand side by finding a common denominator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=num-simp}{2}}{1 - \\cos^2 \\phi} = 18$"},{"latex":"$$\\frac{2}{\\htmlData{anchor=identity}{\\sin^2 \\phi}} = 18$"}],"highlights":[{"color":"sky","style":"box","anchor":"num-simp"},{"color":"sky","style":"box","anchor":"identity"}],"connections":[{"to":"identity","from":"num-simp","color":"sky","label":"1 - \\cos^2\\phi = \\sin^2\\phi","style":"dashed"}]},"order":1,"title":"Simplify the equation","description":"Simplify the numerator by canceling $\\cos \\phi$ and $-\\cos \\phi$. Then use the Pythagorean identity $1 - \\cos^2 \\phi = \\sin^2 \\phi$ in the denominator."},{"type":"text","order":2,"title":"Solve for sine squared","content":"Rearrange the simplified equation to solve for $\\sin^2 \\phi$:\n\n$$2 = 18 \\sin^2 \\phi$$\n\n$$\\sin^2 \\phi = \\frac{2}{18} = \\frac{1}{9}$$"},{"type":"text","order":3,"title":"Solve for cosine squared","content":"Using the Pythagorean identity again, we can find $\\cos^2 \\phi$:\n\n$$\\cos^2 \\phi = 1 - \\sin^2 \\phi$$\n\n$$\\cos^2 \\phi = 1 - \\frac{1}{9} = \\frac{8}{9}$$"},{"type":"text","order":4,"title":"Determine signs in the third quadrant","content":"We are given that $\\phi$ is in the third quadrant, where both sine and cosine are negative:\n\n$$\\sin \\phi = -\\sqrt{\\frac{1}{9}} = -\\frac{1}{3}$$\n\n$$\\cos \\phi = -\\sqrt{\\frac{8}{9}} = -\\frac{\\sqrt{8}}{3} = -\\frac{2\\sqrt{2}}{3}$$"},{"type":"text","order":5,"title":"Calculate cotangent","content":"Using the quotient identity, we calculate $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$:\n\n$$\\cot \\phi = \\frac{-\\frac{2\\sqrt{2}}{3}}{-\\frac{1}{3}} = 2\\sqrt{2}$$\n\nSince both sine and cosine are negative in the third quadrant, their ratio $\\cot \\phi$ is positive."}]},{"id":"reference_triangle","label":"Reference Triangle Method","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"We start by combining the two algebraic fractions on the left-hand side over a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nMultiply the numerators and denominators to get a common denominator of $(1 - \\cos \\phi)(1 + \\cos \\phi)$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$\n\nSimplify the numerator and expand the denominator using the difference of squares:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 18$$"},{"type":"text","order":1,"title":"Find sine of phi","content":"Recall the fundamental trigonometric identity:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies 1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstituting this into our simplified equation:\n\n$$\\frac{2}{\\sin^2 \\phi} = 18$$\n\nRearranging to solve for $\\sin^2 \\phi$:\n\n$$\\sin^2 \\phi = \\frac{2}{18} = \\frac{1}{9}$$\n\nTaking the square root of both sides:\n\n$$\\sin \\phi = \\pm\\frac{1}{3}$$"},{"type":"text","order":2,"title":"Determine sign using quadrant","content":"The question specifies that $\\phi$ lies in the **third quadrant** ($180^\\circ < \\phi < 270^\\circ$).\n\nIn the third quadrant, the sine function is negative ($y < 0$). Therefore, we choose the negative value:\n\n$$\\sin \\phi = -\\frac{1}{3}$$"},{"type":"text","order":3,"title":"Sketch reference triangle","content":"We can represent $\\sin \\phi$ using a reference right-angled triangle in the third quadrant on a Cartesian plane:\n\n$$\\sin \\phi = \\frac{\\text{Opposite}}{\\text{Hypotenuse}} = \\frac{-1}{3}$$\n\n- **Opposite side ($y$):** $-1$ (directed downwards)\n- **Hypotenuse ($r$):** $3$ (always positive)\n\nWe find the **adjacent side ($x$)** using the Pythagorean theorem:\n\n$$x^2 + (-1)^2 = 3^2$$\n$$x^2 + 1 = 9 \\implies x^2 = 8 \\implies x = \\pm\\sqrt{8}$$\n\nSince we are in the third quadrant, the adjacent side must be negative ($x < 0$):\n\n$$x = -\\sqrt{8} = -2\\sqrt{2}$$"},{"type":"text","order":4,"title":"Calculate cotangent","content":"Now we determine the exact value of $\\cot \\phi$ using our reference triangle coordinates:\n\n$$\\cot \\phi = \\frac{\\text{Adjacent}}{\\text{Opposite}} = \\frac{x}{y}$$\n\nSubstitute the values we found:\n\n$$\\cot \\phi = \\frac{-2\\sqrt{2}}{-1} = 2\\sqrt{2}$$\n\nSince both the adjacent and opposite sides are negative in the third quadrant, the cotangent value is positive, as expected."}]}],"generatedAt":"2026-06-09T03:50:52.764Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"73d6108b-c695-4a31-bf42-dc6d0bc69335:lhs_to_rhs","label":"(a) LHS to RHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we start with the expression on the Left-Hand Side (LHS) of the identity:\n\n$$\\text{LHS} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{LHS} = \\frac{\\htmlData{anchor=num1}{\\color{@sky}{1(1 + \\cos \\phi)}} + \\htmlData{anchor=num2}{\\color{@pink}{1(1 - \\cos \\phi)}}}{\\htmlData{anchor=denom}{\\color{@purple}{(1 - \\cos \\phi)(1 + \\cos \\phi)}}}$$"}],"highlights":[{"color":"sky","style":"underline","anchor":"num1"},{"color":"pink","style":"underline","anchor":"num2"},{"color":"purple","style":"underline","anchor":"denom"}],"connections":[]},"order":1,"title":"Find a common denominator","description":"To combine these two algebraic fractions, find a common denominator by multiplying the denominators: $(1 - \\cos \\phi)(1 + \\cos \\phi)$. This earns the first mark."},{"type":"text","order":2,"title":"Simplify the numerator","content":"Now, we expand and simplify the terms in the numerator:\n\n$$(1 + \\cos \\phi) + (1 - \\cos \\phi) = 1 + 1 + \\cos \\phi - \\cos \\phi = 2$$\n\nSo the fraction becomes:\n\n$$\\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$(1 - \\cos \\phi)(1 + \\cos \\phi) = 1^2 - \\cos^2 \\phi$$"},{"latex":"$$$\\text{LHS} = \\frac{2}{\\htmlData{anchor=diff-sq}{\\color{@teal}{1 - \\cos^2 \\phi}}}$$"}],"highlights":[{"color":"teal","style":"box","anchor":"diff-sq"}],"connections":[]},"order":3,"title":"Expand the denominator","description":"We expand the denominator using the difference of two squares identity, $(a-b)(a+b) = a^2 - b^2$, which simplifies our expression to a single term. This gets us the second mark."},{"type":"text","order":4,"title":"Apply Pythagorean identity","content":"Recall the fundamental Pythagorean trigonometric identity from our syllabus:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1$$\n\nRearranging this gives:\n\n$$1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstituting this into our denominator, we get:\n\n$$\\frac{2}{\\sin^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\csc \\phi = \\frac{1}{\\sin \\phi} \\implies \\htmlData{anchor=recip-id}{\\color{@amber}{\\csc^2 \\phi}} = \\htmlData{anchor=sin-id}{\\color{@amber}{\\frac{1}{\\sin^2 \\phi}}}$$"},{"latex":"$$$\\text{LHS} = 2 \\left( \\color{@amber}{\\frac{1}{\\sin^2 \\phi}} \\right) = 2\\color{@amber}{\\csc^2 \\phi} = \\text{RHS}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sin-id"},{"color":"amber","style":"text-color","anchor":"recip-id"}],"connections":[{"to":"recip-id","from":"sin-id","color":"amber","label":null,"style":"solid"}]},"order":5,"title":"Use reciprocal identity","description":"Since the cosecant function is defined as the reciprocal of sine, we can replace the sine term in the denominator. This complete logical step awards the final mark."}]},{"id":"73d6108b-c695-4a31-bf42-dc6d0bc69335:rhs_to_lhs","label":"(a) RHS to LHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using the RHS to LHS manipulation approach, we begin with the right-hand side of the identity:\n\n$$\\text{RHS} = 2 \\csc^2 \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = 2 \\htmlData{anchor=cosec}{\\csc^2 \\phi}$"},{"text":"We use the fundamental reciprocal identity:"},{"latex":"$$\\htmlData{anchor=recip-def}{\\csc \\phi = \\frac{1}{\\sin \\phi}}$"},{"text":"Substituting this in gives:"},{"latex":"$$\\phantom{\\text{RHS}} = \\htmlData{anchor=sin-sub}{\\frac{2}{\\sin^2 \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cosec"},{"color":"sky","style":"text-color","anchor":"recip-def"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"recip-def","from":"cosec","color":"sky","label":"reciprocal","style":"solid"},{"to":"sin-sub","from":"recip-def","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Convert cosecant to sine","description":"Convert the cosecant term to its equivalent expression in terms of sine."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2}{\\htmlData{anchor=sin-denom}{\\sin^2 \\phi}}$"},{"text":"Using the Pythagorean identity:"},{"latex":"$$\\htmlData{anchor=pythag}{\\sin^2 \\phi + \\cos^2 \\phi = 1} \\implies \\htmlData{anchor=sin-isolated}{\\sin^2 \\phi = 1 - \\cos^2 \\phi}$"},{"text":"We substitute this into our denominator:"},{"latex":"$$\\phantom{\\text{RHS}} = \\frac{2}{\\htmlData{anchor=cos-denom}{1 - \\cos^2 \\phi}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sin-denom"},{"color":"amber","style":"text-color","anchor":"pythag"},{"color":"amber","style":"text-color","anchor":"cos-denom"}],"connections":[{"to":"pythag","from":"sin-denom","color":"amber","label":"identity","style":"solid"},{"to":"cos-denom","from":"sin-isolated","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply Pythagorean identity","description":"Rewrite the denominator using the Pythagorean identity."},{"type":"text","order":3,"title":"Factor the denominator","content":"The expression in the denominator, $1 - \\cos^2 \\phi$, is a difference of two squares. We can factor it as:\n\n$$1 - \\cos^2 \\phi = (1 - \\cos \\phi)(1 + \\cos \\phi)$$\n\nThis updates our fraction to:\n\n$$\\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":4,"title":"Split the fraction","content":"To rewrite this single fraction as two separate fractions, we decompose the numerator. Notice that we can write $2$ as:\n\n$$2 = (1 + \\cos \\phi) + (1 - \\cos \\phi)$$\n\nNow we substitute this into the numerator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nWe can now split this into two separate terms:\n\n$$\\frac{1 + \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)} + \\frac{1 - \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":5,"title":"Simplify to LHS","content":"By cancelling the common factors in the numerator and denominator of each fraction, we get:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = \\text{LHS}$$\n\nSince we have shown that $\\text{RHS} \\equiv \\text{LHS}$, the identity is proven."}]},{"id":"297bfcd9-44d4-443c-b4e0-8c24a73e27a2:trigonometric_simplification","label":"(b) Trigonometric Simplification","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"To solve for $k$, we start by combining the two fractions on the left-hand side under a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = k$$\n\nMultiply the numerator and denominator of each fraction to get a common denominator of $(1 - \\cos \\phi)(1 + \\cos \\phi)$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = k$$"},{"type":"text","order":1,"title":"Use trigonometric identity","content":"Simplify the numerator and expand the denominator:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = k$$\n\nUsing the fundamental Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$, we can substitute $1 - \\cos^2 \\phi = \\sin^2 \\phi$ in the denominator:\n\n$$k = \\frac{2}{\\sin^2 \\phi}$$\n\nThis earns you the first mark (**M1**) for simplifying the relation."},{"type":"text","order":2,"title":"Determine domain restrictions","content":"For the original expression to be defined, the denominators cannot be zero:\n\n$$1 - \\cos \\phi \\neq 0 \\implies \\cos \\phi \\neq 1$$\n$$1 + \\cos \\phi \\neq 0 \\implies \\cos \\phi \\neq -1$$\n\nSince $\\cos \\phi \\neq \\pm 1$, it follows that $\\sin \\phi \\neq 0$.\n\nKnowing that for any real angle $\\phi$, the range of the sine function is $-1 \\le \\sin \\phi \\le 1$, squaring both sides gives:\n\n$$0 \\le \\sin^2 \\phi \\le 1$$\n\nBecause $\\sin \\phi \\neq 0$, this simplifies to the restricted range:\n\n$$0 < \\sin^2 \\phi \\le 1$$\n\nThis justification earns you the second mark (**M1**)."},{"type":"text","order":3,"title":"Find the range of k","content":"Now we use the inequality $0 < \\sin^2 \\phi \\le 1$ to find the range of $k$:\n\nTaking the reciprocal of both sides reverses the inequality sign:\n\n$$\\frac{1}{\\sin^2 \\phi} \\ge 1$$\n\nMultiplying by $2$:\n\n$$k = \\frac{2}{\\sin^2 \\phi} \\ge 2$$\n\nThus, the set of values of $k$ for which the equation has real solutions is:\n\n$$k \\ge 2$$\n\nThis final conclusion earns you the final accuracy mark (**A1**)."}]},{"id":"297bfcd9-44d4-443c-b4e0-8c24a73e27a2:algebraic_substitution","label":"(b) Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Introduce algebraic substitution","content":"To solve this problem using algebraic substitution, we define a new variable for the trigonometric term. Let:\n\n$$u = \\cos \\phi$$\n\nFor the fractions in the equation to be defined, the denominators cannot be zero. Therefore, $\\cos \\phi \\neq 1$ and $\\cos \\phi \\neq -1$.\n\nSince the range of the cosine function is $-1 \\le \\cos \\phi \\le 1$, excluding the boundary values gives the domain constraint for $u$:\n\n$$-1 < u < 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{1}{1 - \\htmlData{anchor=u-sub1}{\\color{@sky}{u}}} + \\frac{1}{1 + \\htmlData{anchor=u-sub2}{\\color{@sky}{u}}} = k$$"},{"latex":"$$$\\frac{(1 + \\color{@sky}{u}) + (1 - \\color{@sky}{u})}{(1 - \\color{@sky}{u})(1 + \\color{@sky}{u})} = k$$"},{"latex":"$$$\\frac{2}{1 - \\htmlData{anchor=u-sq}{\\color{@purple}{u^2}}} = k$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-sub1"},{"color":"sky","style":"text-color","anchor":"u-sub2"},{"color":"purple","style":"text-color","anchor":"u-sq"}],"connections":[{"to":"u-sq","from":"u-sub1","color":"purple","label":"combine terms","style":"solid"}]},"order":1,"title":"Simplify the expression","description":"Substitute $u$ into the equation and find a common denominator to simplify the algebraic fractions."},{"type":"text","order":2,"title":"Apply domain constraints","content":"Now we analyze the range of the term $u^2$. Since $-1 < u < 1$, squaring both sides yields:\n\n$$0 \\le u^2 < 1$$\n\nWe can build up the expression for $k$ step-by-step using inequalities:\n\n1. Multiply by $-1$:\n$$-1 < -u^2 \\le 0$$\n\n2. Add $1$ to all parts:\n$$0 < 1 - u^2 \\le 1$$\n\n3. Take the reciprocal (reversing the inequality direction for positive values):\n$$\\frac{1}{1 - u^2} \\ge 1$$\n\n4. Multiply by $2$:\n$$\\frac{2}{1 - u^2} \\ge 2$$"},{"type":"text","order":3,"title":"State the final set","content":"Since $k = \\frac{2}{1 - u^2}$, we conclude that:\n\n$$k \\ge 2$$\n\nThis is the complete set of values for which the equation has real solutions for $\\phi$."}]},{"id":"d766397d-e937-4d6e-a926-38652bba6ec1:pythagorean_identity","label":"(c) Using Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Simplify the algebraic fraction","content":"First, we can combine the fractions on the left-hand side of the equation by finding a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nMultiply the numerator and denominator of each fraction to combine them:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$\n\nSimplify the numerator and expand the denominator:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 18$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the fundamental identity: } \\htmlData{anchor=trig-identity}{\\color{@sky}{1 - \\cos^2 \\phi}} = \\htmlData{anchor=sin-sq}{\\color{@sky}{\\sin^2 \\phi}}$"},{"latex":"$${\\frac{2}{\\htmlData{anchor=sub-denom}{\\color{@sky}{1 - \\cos^2 \\phi}}}} = 18 \\implies \\frac{2}{\\htmlData{anchor=sin-denom}{\\color{@sky}{\\sin^2 \\phi}}} = 18$"},{"latex":"$$\\text{Since } \\htmlData{anchor=cosec-def}{\\color{@amber}{\\csc^2 \\phi}} = \\htmlData{anchor=inv-sin}{\\color{@amber}{\\frac{1}{\\sin^2 \\phi}}}, \\text{ we can write:}$"},{"latex":"$$2\\htmlData{anchor=cosec-term}{\\color{@amber}{\\csc^2 \\phi}} = 18$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"trig-identity"},{"color":"sky","style":"text-color","anchor":"sin-sq"},{"color":"amber","style":"text-color","anchor":"cosec-def"}],"connections":[{"to":"sub-denom","from":"trig-identity","color":"sky","label":"substitute","style":"solid"},{"to":"cosec-term","from":"cosec-def","color":"amber","label":"rewrite","style":"solid"}]},"order":1,"title":"Express in terms of cosecant","description":"We use the identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$ to rewrite the denominator, and then express the expression in terms of the cosecant function."},{"type":"text","order":2,"title":"Solve for cosecant squared","content":"Now, we divide both sides by $2$ to solve for $\\csc^2 \\phi$:\n\n$$\\csc^2 \\phi = 9$$\n\n*(This earns the first mark B1)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Recall the Pythagorean identity: } \\htmlData{anchor=pythag-id}{\\color{@purple}{1 + \\cot^2 \\phi}} = \\htmlData{anchor=cosec-sq-term}{\\color{@purple}{\\csc^2 \\phi}}$"},{"latex":"$$1 + \\cot^2 \\phi = \\htmlData{anchor=num-val}{\\color{@purple}{9}}$"},{"latex":"$$\\cot^2 \\phi = 8$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"pythag-id"},{"color":"purple","style":"text-color","anchor":"cosec-sq-term"}],"connections":[{"to":"cosec-sq-term","from":"pythag-id","color":"purple","label":"identity","style":"solid"}]},"order":3,"title":"Apply the Pythagorean identity","description":"Using the Pythagorean identity relating cotangent and cosecant, we substitute $\\csc^2 \\phi = 9$ and solve for $\\cot^2 \\phi$."},{"type":"text","order":4,"title":"Quadrant analysis and final value","content":"Taking the square root of both sides gives:\n\n$$\\cot \\phi = \\pm\\sqrt{8} = \\pm 2\\sqrt{2}$$\n\nWe are given that $\\phi$ is in the **third quadrant** ($180^\\circ < \\phi < 270^\\circ$).\n\nIn the third quadrant:\n- Both $\\sin \\phi$ and $\\cos \\phi$ are negative.\n- Therefore, $\\tan \\phi$ and its reciprocal $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$ must be **positive**.\n\nThus, we select the positive root:\n\n$$\\cot \\phi = 2\\sqrt{2}$$\n\n*(Earning M1 for finding $\\cot^2 \\phi = 8$ and A1 for selecting the positive root with correct quadrant justification)*"}]},{"id":"d766397d-e937-4d6e-a926-38652bba6ec1:sine_cosine_ratio","label":"(c) Finding Sine and Cosine","steps":[{"type":"text","order":0,"title":"Combine the algebraic fractions","content":"To solve the equation\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nwe first combine the fractions on the left-hand side by finding a common denominator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=num-simp}{2}}{1 - \\cos^2 \\phi} = 18$"},{"latex":"$$\\frac{2}{\\htmlData{anchor=identity}{\\sin^2 \\phi}} = 18$"}],"highlights":[{"color":"sky","style":"box","anchor":"num-simp"},{"color":"sky","style":"box","anchor":"identity"}],"connections":[{"to":"identity","from":"num-simp","color":"sky","label":"1 - \\cos^2\\phi = \\sin^2\\phi","style":"dashed"}]},"order":1,"title":"Simplify the equation","description":"Simplify the numerator by canceling $\\cos \\phi$ and $-\\cos \\phi$. Then use the Pythagorean identity $1 - \\cos^2 \\phi = \\sin^2 \\phi$ in the denominator."},{"type":"text","order":2,"title":"Solve for sine squared","content":"Rearrange the simplified equation to solve for $\\sin^2 \\phi$:\n\n$$2 = 18 \\sin^2 \\phi$$\n\n$$\\sin^2 \\phi = \\frac{2}{18} = \\frac{1}{9}$$"},{"type":"text","order":3,"title":"Solve for cosine squared","content":"Using the Pythagorean identity again, we can find $\\cos^2 \\phi$:\n\n$$\\cos^2 \\phi = 1 - \\sin^2 \\phi$$\n\n$$\\cos^2 \\phi = 1 - \\frac{1}{9} = \\frac{8}{9}$$"},{"type":"text","order":4,"title":"Determine signs in the third quadrant","content":"We are given that $\\phi$ is in the third quadrant, where both sine and cosine are negative:\n\n$$\\sin \\phi = -\\sqrt{\\frac{1}{9}} = -\\frac{1}{3}$$\n\n$$\\cos \\phi = -\\sqrt{\\frac{8}{9}} = -\\frac{\\sqrt{8}}{3} = -\\frac{2\\sqrt{2}}{3}$$"},{"type":"text","order":5,"title":"Calculate cotangent","content":"Using the quotient identity, we calculate $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$:\n\n$$\\cot \\phi = \\frac{-\\frac{2\\sqrt{2}}{3}}{-\\frac{1}{3}} = 2\\sqrt{2}$$\n\nSince both sine and cosine are negative in the third quadrant, their ratio $\\cot \\phi$ is positive."}]},{"id":"d766397d-e937-4d6e-a926-38652bba6ec1:reference_triangle","label":"(c) Reference Triangle Method","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"We start by combining the two algebraic fractions on the left-hand side over a common denominator:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 18$$\n\nMultiply the numerators and denominators to get a common denominator of $(1 - \\cos \\phi)(1 + \\cos \\phi)$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 18$$\n\nSimplify the numerator and expand the denominator using the difference of squares:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 18$$"},{"type":"text","order":1,"title":"Find sine of phi","content":"Recall the fundamental trigonometric identity:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies 1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstituting this into our simplified equation:\n\n$$\\frac{2}{\\sin^2 \\phi} = 18$$\n\nRearranging to solve for $\\sin^2 \\phi$:\n\n$$\\sin^2 \\phi = \\frac{2}{18} = \\frac{1}{9}$$\n\nTaking the square root of both sides:\n\n$$\\sin \\phi = \\pm\\frac{1}{3}$$"},{"type":"text","order":2,"title":"Determine sign using quadrant","content":"The question specifies that $\\phi$ lies in the **third quadrant** ($180^\\circ < \\phi < 270^\\circ$).\n\nIn the third quadrant, the sine function is negative ($y < 0$). Therefore, we choose the negative value:\n\n$$\\sin \\phi = -\\frac{1}{3}$$"},{"type":"text","order":3,"title":"Sketch reference triangle","content":"We can represent $\\sin \\phi$ using a reference right-angled triangle in the third quadrant on a Cartesian plane:\n\n$$\\sin \\phi = \\frac{\\text{Opposite}}{\\text{Hypotenuse}} = \\frac{-1}{3}$$\n\n- **Opposite side ($y$):** $-1$ (directed downwards)\n- **Hypotenuse ($r$):** $3$ (always positive)\n\nWe find the **adjacent side ($x$)** using the Pythagorean theorem:\n\n$$x^2 + (-1)^2 = 3^2$$\n$$x^2 + 1 = 9 \\implies x^2 = 8 \\implies x = \\pm\\sqrt{8}$$\n\nSince we are in the third quadrant, the adjacent side must be negative ($x < 0$):\n\n$$x = -\\sqrt{8} = -2\\sqrt{2}$$"},{"type":"text","order":4,"title":"Calculate cotangent","content":"Now we determine the exact value of $\\cot \\phi$ using our reference triangle coordinates:\n\n$$\\cot \\phi = \\frac{\\text{Adjacent}}{\\text{Opposite}} = \\frac{x}{y}$$\n\nSubstitute the values we found:\n\n$$\\cot \\phi = \\frac{-2\\sqrt{2}}{-1} = 2\\sqrt{2}$$\n\nSince both the adjacent and opposite sides are negative in the third quadrant, the cotangent value is positive, as expected."}]}],"generatedAt":"2026-06-09T03:51:09.599Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"2300c62d-9b78-42e6-acf9-1b92a50bd19f","revisionId":"be9a2922-5123-4090-8115-cace97a77259","questionNumber":null,"questionType":"LA","specification":"The minute hand of a clock is $12$ cm long. Find, in exact form,","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"47a7eea2-7423-4f95-b13a-2bd39f914619","label":"a","content":"the distance travelled by its tip in $35$ minutes;","markscheme":"- Angle swept $=\\frac{35}{60}(2\\pi)=\\frac{7\\pi}{6}$ radians M1\n- Uses $s=r\\theta$ M1\n- Distance $=12\\times\\frac{7\\pi}{6}=14\\pi$ cm A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"23798786-fa26-4785-8da0-e6475bd60635","label":"b","content":"the area swept out by the hand in that time.","markscheme":"- $A=\\frac12(12^2)\\frac{7\\pi}{6}$ M1\n- $A=84\\pi$ cm$^2$ A1","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"2423687b-ba39-452d-8863-34deb93dac6c","revisionId":"8fd25f2e-fa8f-4057-bc6e-0cd83e389438","questionNumber":null,"questionType":"LA","specification":"The efficiency $E$ of a precision-engineered joint in a robotic limb is determined by the tilt angle $\\phi$ such that\n\n\\[ (\\sin \\phi + 12 \\cos \\phi)^2 + (4 \\sin \\phi - 3 \\cos \\phi)^2 = 119 \\]","difficulty":4,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"1d595347-ea75-479f-81c0-e99fdac05ba6","label":null,"content":"Given that the joint is positioned at an obtuse angle, such that $\\frac{\\pi}{2} < \\phi < \\pi$, determine the exact value of $\\sin \\phi$. Fully justify your answer.","markscheme":"- Expands the first bracket to obtain $\\sin^2 \\phi + 24 \\sin \\phi \\cos \\phi + 144 \\cos^2 \\phi$ M1\n- Expands the second bracket to obtain $16 \\sin^2 \\phi - 24 \\sin \\phi \\cos \\phi + 9 \\cos^2 \\phi$ A1\n- Combines terms to reach the simplified equation $17 \\sin^2 \\phi + 153 \\cos^2 \\phi = 119$ A1\n- Substitutes $\\cos^2 \\phi = 1 - \\sin^2 \\phi$ into the equation: $17 \\sin^2 \\phi + 153(1 - \\sin^2 \\phi) = 119$ M1\n- Simplifies to $-136 \\sin^2 \\phi = -34$, yielding $\\sin^2 \\phi = \\frac{1}{4}$ A1\n- States that $\\sin \\phi = \\frac{1}{2}$ and justifies the positive root by stating that for $\\frac{\\pi}{2} < \\phi < \\pi$ (second quadrant), the sine function is positive A1","marks":6,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"sine_identity","label":"Using Pythagorean Identity for Sine","steps":[{"type":"text","order":0,"title":"Expand the first term","content":"To solve the equation, we first expand the first squared term, $(\text{sin } \nu + 12 \text{cos } \nu)^2$, using the identity $(a+b)^2 = a^2 + 2ab + b^2$:\n\n$$(\\sin \\phi + 12 \\cos \\phi)^2 = \\sin^2 \\phi + 24 \\sin \\phi \\cos \\phi + 144 \\cos^2 \\phi$$","highlights":[{"text":"(\\sin \\phi + 12 \\cos \\phi)^2 + (4 \\sin \\phi - 3 \\cos \\phi)^2 = 119","location":"specification"}]},{"type":"text","order":1,"title":"Expand the second term","content":"Next, we expand the second squared term, $(4 \\sin \\phi - 3 \\cos \\phi)^2$, using the identity $(a-b)^2 = a^2 - 2ab + b^2$:\n\n$$(4 \\sin \\phi - 3 \\cos \\phi)^2 = 16 \\sin^2 \\phi - 24 \\sin \\phi \\cos \\phi + 9 \\cos^2 \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$(\\sin^2 \\phi + \\htmlData{anchor=cross-pos}{\\color{@sky}{24 \\sin \\phi \\cos \\phi}} + 144 \\cos^2 \\phi) + (16 \\sin^2 \\phi \\htmlData{anchor=cross-neg}{\\color{@sky}{- 24 \\sin \\phi \\cos \\phi}} + 9 \\cos^2 \\phi) = 119$"},{"latex":"$$\\htmlData{anchor=simp-eq}{17 \\sin^2 \\phi + 153 \\cos^2 \\phi = 119}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cross-pos"},{"color":"sky","style":"text-color","anchor":"cross-neg"}],"connections":[{"to":"cross-neg","from":"cross-pos","color":"sky","label":"cancel out","style":"dashed"}]},"order":2,"title":"Combine and simplify","description":"Now, we add the two expanded expressions together. Notice that the cross-terms cancel out: $+24 \\sin \\phi \\cos \\phi - 24 \\sin \\phi \\cos \\phi = 0$. Grouping the remaining like terms gives the simplified equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$17 \\sin^2 \\phi + 153 \\htmlData{anchor=cos-term}{\\color{@amber}{\\cos^2 \\phi}} = 119$$"},{"latex":"$$$17 \\sin^2 \\phi + 153\\htmlData{anchor=cos-sub}{\\color{@amber}{(1 - \\sin^2 \\phi)}} = 119$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"cos-term"},{"color":"amber","style":"text-color","anchor":"cos-sub"}],"connections":[{"to":"cos-sub","from":"cos-term","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Apply Pythagorean identity","description":"To solve for $\\sin \\phi$, we use the fundamental Pythagorean identity: $$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies \\cos^2 \\phi = 1 - \\sin^2 \\phi$$ Substitute this into our simplified equation."},{"type":"text","order":4,"title":"Solve for sine squared","content":"Now, expand the bracket and collect the $\\sin^2 \\phi$ terms:\n\n$$\\begin{aligned} 17 \\sin^2 \\phi + 153 - 153 \\sin^2 \\phi &= 119 \\\\ -136 \\sin^2 \\phi + 153 &= 119 \\\\ -136 \\sin^2 \\phi &= -34 \\\\ \\sin^2 \\phi &= \\frac{-34}{-136} \\\\ \\sin^2 \\phi &= \\frac{1}{4} \\end{aligned}$$"},{"type":"text","order":5,"title":"Determine the final value","content":"Taking the square root of both sides gives:\n\n$$\\sin \\phi = \\pm\\frac{1}{2}$$\n\nWe are given that the joint is at an obtuse angle, specifically in the second quadrant where:\n\n$$\\frac{\\pi}{2} < \\phi < \\pi$$\n\nIn the second quadrant, the sine function is positive (as verified by the ASTC / CAST diagram). Therefore, we select the positive root:\n\n$$\\sin \\phi = \\frac{1}{2}$$","highlights":[{"text":"obtuse angle","location":"specification"},{"text":"\\frac{\\pi}{2} < \\phi < \\pi","location":"specification"}]}]},{"id":"tangent_substitution","label":"Solving via Tangent Identity","steps":[{"type":"text","order":0,"title":"Expand the terms","content":"First, let's expand both squared terms in the given equation:\n\n$$(\\sin \\phi + 12 \\cos \\phi)^2 + (4 \\sin \\phi - 3 \\cos \\phi)^2 = 119$$\n\nExpanding the first bracket:\n$$(\\sin \\phi + 12 \\cos \\phi)^2 = \\sin^2 \\phi + 24 \\sin \\phi \\cos \\phi + 144 \\cos^2 \\phi$$\n\nExpanding the second bracket:\n$$(4 \\sin \\phi - 3 \\cos \\phi)^2 = 16 \\sin^2 \\phi - 24 \\sin \\phi \\cos \\phi + 9 \\cos^2 \\phi$$"},{"type":"text","order":1,"title":"Simplify the equation","content":"Now we sum these two expansions together. Notice that the cross terms $+24 \\sin \\phi \\cos \\phi$ and $-24 \\sin \\phi \\cos \\phi$ cancel each other out:\n\n$$\\begin{aligned}\n&(\\sin^2 \\phi + 16 \\sin^2 \\phi) + (144 \\cos^2 \\phi + 9 \\cos^2 \\phi) = 119 \\\\\n&17 \\sin^2 \\phi + 153 \\cos^2 \\phi = 119\n\\end{aligned}$$"},{"type":"text","order":2,"title":"Convert to tangent","content":"To solve this using the tangent identity, we divide every term in the equation by $\\cos^2 \\phi$ (assuming $\\cos \\phi \\neq 0$):\n\n$$\\frac{17 \\sin^2 \\phi}{\\cos^2 \\phi} + \\frac{153 \\cos^2 \\phi}{\\cos^2 \\phi} = \\frac{119}{\\cos^2 \\phi}$$\n\nUsing the identities $\\tan \\phi = \\frac{\\sin \\phi}{\\cos \\phi}$ and $\\sec^2 \\phi = \\frac{1}{\\cos^2 \\phi} = 1 + \\tan^2 \\phi$, we get:\n\n$$17 \\tan^2 \\phi + 153 = 119 (1 + \\tan^2 \\phi)$$"},{"type":"text","order":3,"title":"Solve for tangent","content":"Now, let's solve for $\\tan^2 \\phi$:\n\n$$\\begin{aligned}\n17 \\tan^2 \\phi + 153 &= 119 + 119 \\tan^2 \\phi \\\\\n153 - 119 &= 119 \\tan^2 \\phi - 17 \\tan^2 \\phi \\\\\n34 &= 102 \\tan^2 \\phi \\\\\n\\tan^2 \\phi &= \\frac{34}{102} = \\frac{1}{3}\n\\end{aligned}$$"},{"type":"text","order":4,"title":"Determine the tangent sign","content":"We are given that the joint is positioned at an obtuse angle in the second quadrant, where $\\frac{\\pi}{2} < \\phi < \\pi$.\n\nIn the second quadrant, tangent is negative. Therefore, taking the square root of both sides, we get:\n\n$$\\tan \\phi = -\\frac{1}{\\sqrt{3}}$$","highlights":[{"text":"the joint is positioned at an obtuse angle, such that $\\frac{\\pi}{2} < \\phi < \\pi$","location":"specification"}]},{"type":"text","order":5,"title":"Find the sine value","content":"Using the relationship between trig ratios, if $\\tan \\phi = -\\frac{1}{\\sqrt{3}}$, we can construct a right-angled triangle with an opposite side of $1$ and an adjacent side of $\\sqrt{3}$.\n\nThe hypotenuse is:\n$$\\text{Hypotenuse} = \\sqrt{1^2 + (\\sqrt{3})^2} = \\sqrt{1 + 3} = 2$$\n\nThus, the magnitude of $\\sin \\phi$ is:\n$$|\\sin \\phi| = \\frac{\\text{Opposite}}{\\text{Hypotenuse}} = \\frac{1}{2}$$\n\nSince $\\phi$ is in the second quadrant ($\\frac{\\pi}{2} < \\phi < \\pi$), $\\sin \\phi$ must be positive. Therefore:\n\n$$\\sin \\phi = \\frac{1}{2}$$","highlights":[{"text":"determine the exact value of $\\sin \\phi$","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:39:30.745Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"1d595347-ea75-479f-81c0-e99fdac05ba6:sine_identity","label":"Using Pythagorean Identity for Sine","steps":[{"type":"text","order":0,"title":"Expand the first term","content":"To solve the equation, we first expand the first squared term, $(\\text{sin } \nu + 12 \\text{cos } \nu)^2$, using the identity $(a+b)^2 = a^2 + 2ab + b^2$:\n\n$$(\\sin \\phi + 12 \\cos \\phi)^2 = \\sin^2 \\phi + 24 \\sin \\phi \\cos \\phi + 144 \\cos^2 \\phi$$","highlights":[{"text":"(\\sin \\phi + 12 \\cos \\phi)^2 + (4 \\sin \\phi - 3 \\cos \\phi)^2 = 119","location":"specification"}]},{"type":"text","order":1,"title":"Expand the second term","content":"Next, we expand the second squared term, $(4 \\sin \\phi - 3 \\cos \\phi)^2$, using the identity $(a-b)^2 = a^2 - 2ab + b^2$:\n\n$$(4 \\sin \\phi - 3 \\cos \\phi)^2 = 16 \\sin^2 \\phi - 24 \\sin \\phi \\cos \\phi + 9 \\cos^2 \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$(\\sin^2 \\phi + \\htmlData{anchor=cross-pos}{\\color{@sky}{24 \\sin \\phi \\cos \\phi}} + 144 \\cos^2 \\phi) + (16 \\sin^2 \\phi \\htmlData{anchor=cross-neg}{\\color{@sky}{- 24 \\sin \\phi \\cos \\phi}} + 9 \\cos^2 \\phi) = 119$"},{"latex":"$$\\htmlData{anchor=simp-eq}{17 \\sin^2 \\phi + 153 \\cos^2 \\phi = 119}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cross-pos"},{"color":"sky","style":"text-color","anchor":"cross-neg"}],"connections":[{"to":"cross-neg","from":"cross-pos","color":"sky","label":"cancel out","style":"dashed"}]},"order":2,"title":"Combine and simplify","description":"Now, we add the two expanded expressions together. Notice that the cross-terms cancel out: $+24 \\sin \\phi \\cos \\phi - 24 \\sin \\phi \\cos \\phi = 0$. Grouping the remaining like terms gives the simplified equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$17 \\sin^2 \\phi + 153 \\htmlData{anchor=cos-term}{\\color{@amber}{\\cos^2 \\phi}} = 119$$"},{"latex":"$$$17 \\sin^2 \\phi + 153\\htmlData{anchor=cos-sub}{\\color{@amber}{(1 - \\sin^2 \\phi)}} = 119$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"cos-term"},{"color":"amber","style":"text-color","anchor":"cos-sub"}],"connections":[{"to":"cos-sub","from":"cos-term","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Apply Pythagorean identity","description":"To solve for $\\sin \\phi$, we use the fundamental Pythagorean identity: $$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies \\cos^2 \\phi = 1 - \\sin^2 \\phi$$ Substitute this into our simplified equation."},{"type":"text","order":4,"title":"Solve for sine squared","content":"Now, expand the bracket and collect the $\\sin^2 \\phi$ terms:\n\n$$\\begin{aligned} 17 \\sin^2 \\phi + 153 - 153 \\sin^2 \\phi &= 119 \\\\ -136 \\sin^2 \\phi + 153 &= 119 \\\\ -136 \\sin^2 \\phi &= -34 \\\\ \\sin^2 \\phi &= \\frac{-34}{-136} \\\\ \\sin^2 \\phi &= \\frac{1}{4} \\end{aligned}$$"},{"type":"text","order":5,"title":"Determine the final value","content":"Taking the square root of both sides gives:\n\n$$\\sin \\phi = \\pm\\frac{1}{2}$$\n\nWe are given that the joint is at an obtuse angle, specifically in the second quadrant where:\n\n$$\\frac{\\pi}{2} < \\phi < \\pi$$\n\nIn the second quadrant, the sine function is positive (as verified by the ASTC / CAST diagram). Therefore, we select the positive root:\n\n$$\\sin \\phi = \\frac{1}{2}$$","highlights":[{"text":"obtuse angle","location":"specification"},{"text":"\\frac{\\pi}{2} < \\phi < \\pi","location":"specification"}]}]},{"id":"1d595347-ea75-479f-81c0-e99fdac05ba6:tangent_substitution","label":"Solving via Tangent Identity","steps":[{"type":"text","order":0,"title":"Expand the terms","content":"First, let's expand both squared terms in the given equation:\n\n$$(\\sin \\phi + 12 \\cos \\phi)^2 + (4 \\sin \\phi - 3 \\cos \\phi)^2 = 119$$\n\nExpanding the first bracket:\n$$(\\sin \\phi + 12 \\cos \\phi)^2 = \\sin^2 \\phi + 24 \\sin \\phi \\cos \\phi + 144 \\cos^2 \\phi$$\n\nExpanding the second bracket:\n$$(4 \\sin \\phi - 3 \\cos \\phi)^2 = 16 \\sin^2 \\phi - 24 \\sin \\phi \\cos \\phi + 9 \\cos^2 \\phi$$"},{"type":"text","order":1,"title":"Simplify the equation","content":"Now we sum these two expansions together. Notice that the cross terms $+24 \\sin \\phi \\cos \\phi$ and $-24 \\sin \\phi \\cos \\phi$ cancel each other out:\n\n$$\\begin{aligned}\n&(\\sin^2 \\phi + 16 \\sin^2 \\phi) + (144 \\cos^2 \\phi + 9 \\cos^2 \\phi) = 119 \\\\\n&17 \\sin^2 \\phi + 153 \\cos^2 \\phi = 119\n\\end{aligned}$$"},{"type":"text","order":2,"title":"Convert to tangent","content":"To solve this using the tangent identity, we divide every term in the equation by $\\cos^2 \\phi$ (assuming $\\cos \\phi \\neq 0$):\n\n$$\\frac{17 \\sin^2 \\phi}{\\cos^2 \\phi} + \\frac{153 \\cos^2 \\phi}{\\cos^2 \\phi} = \\frac{119}{\\cos^2 \\phi}$$\n\nUsing the identities $\\tan \\phi = \\frac{\\sin \\phi}{\\cos \\phi}$ and $\\sec^2 \\phi = \\frac{1}{\\cos^2 \\phi} = 1 + \\tan^2 \\phi$, we get:\n\n$$17 \\tan^2 \\phi + 153 = 119 (1 + \\tan^2 \\phi)$$"},{"type":"text","order":3,"title":"Solve for tangent","content":"Now, let's solve for $\\tan^2 \\phi$:\n\n$$\\begin{aligned}\n17 \\tan^2 \\phi + 153 &= 119 + 119 \\tan^2 \\phi \\\\\n153 - 119 &= 119 \\tan^2 \\phi - 17 \\tan^2 \\phi \\\\\n34 &= 102 \\tan^2 \\phi \\\\\n\\tan^2 \\phi &= \\frac{34}{102} = \\frac{1}{3}\n\\end{aligned}$$"},{"type":"text","order":4,"title":"Determine the tangent sign","content":"We are given that the joint is positioned at an obtuse angle in the second quadrant, where $\\frac{\\pi}{2} < \\phi < \\pi$.\n\nIn the second quadrant, tangent is negative. Therefore, taking the square root of both sides, we get:\n\n$$\\tan \\phi = -\\frac{1}{\\sqrt{3}}$$","highlights":[{"text":"the joint is positioned at an obtuse angle, such that $\\frac{\\pi}{2} < \\phi < \\pi$","location":"specification"}]},{"type":"text","order":5,"title":"Find the sine value","content":"Using the relationship between trig ratios, if $\\tan \\phi = -\\frac{1}{\\sqrt{3}}$, we can construct a right-angled triangle with an opposite side of $1$ and an adjacent side of $\\sqrt{3}$.\n\nThe hypotenuse is:\n$$\\text{Hypotenuse} = \\sqrt{1^2 + (\\sqrt{3})^2} = \\sqrt{1 + 3} = 2$$\n\nThus, the magnitude of $\\sin \\phi$ is:\n$$|\\sin \\phi| = \\frac{\\text{Opposite}}{\\text{Hypotenuse}} = \\frac{1}{2}$$\n\nSince $\\phi$ is in the second quadrant ($\\frac{\\pi}{2} < \\phi < \\pi$), $\\sin \\phi$ must be positive. Therefore:\n\n$$\\sin \\phi = \\frac{1}{2}$$","highlights":[{"text":"determine the exact value of $\\sin 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Trigonometry (A-level only)","topicId":"97cb7c2e-f32f-45ff-ada1-651a91e4cf22","topicSidebar":"$L3c","topicSlug":"trigonometry-8","topicTitle":"3.4 Trigonometry (A-level only)"}]}],"$L3d","$L3e"]}]

3.4 Trigonometry (A-level only)

Question 67

3.4 Trigonometry (A-level only) Questions

3.4 Trigonometry (A-level only)

Question 67

3.4 Trigonometry (A-level only) Questions