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\\cos^2 \\phi}$ M1\n- Uses the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$ to reach $\\frac{2}{\\sin^2 \\phi} = 2 \\csc^2 \\phi$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"lhs_to_rhs","label":"Starting from LHS","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we start with the Left-Hand Side (LHS) of the equation:\n\n$$\\text{LHS} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$"},{"type":"text","order":1,"title":"Combine the fractions","content":"To combine the two algebraic fractions, we find a common denominator by multiplying the two denominators: $(1 - \\cos \\phi)(1 + \\cos \\phi)$.\n\nWe rewrite each fraction with this common denominator:\n\n$$\\frac{1 \\cdot (1 + \\cos \\phi) + 1 \\cdot (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nThis step earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Simplify numerator and denominator","content":"Now, let's simplify both the numerator and the denominator:\n\n1. **Numerator:** Combine the terms in the numerator:\n $$(1 + \\cos \\phi) + (1 - \\cos \\phi) = 2$$\n\n2. **Denominator:** Expand using the difference of two squares identity, $(a-b)(a+b) = a^2 - b^2$:\n $$(1 - \\cos \\phi)(1 + \\cos \\phi) = 1^2 - \\cos^2 \\phi = 1 - \\cos^2 \\phi$$\n\nSubstituting these back into our expression gives:\n\n$$\\frac{2}{1 - \\cos^2 \\phi}$$\n\nSimplifying to this form earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Apply trigonometric identities","content":"Recall the fundamental Pythagorean identity:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies 1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstitute this into the denominator:\n\n$$\\frac{2}{\\sin^2 \\phi}$$\n\nNext, use the reciprocal trigonometric identity for cosecant, $\\csc \\phi = \\frac{1}{\\sin \\phi}$, which also means $\\csc^2 \\phi = \\frac{1}{\\sin^2 \\phi}$:\n\n$$\\frac{2}{\\sin^2 \\phi} = 2 \\left(\\frac{1}{\\sin^2 \\phi}\\right) = 2 \\csc^2 \\phi$$\n\nThis is equal to the Right-Hand Side (RHS). Showing this substitution clearly to reach the final expression completes the proof and earns the final accuracy mark (**A1**)."}]},{"id":"rhs_to_lhs","label":"Starting from RHS","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity by starting from the Right-Hand Side (RHS), we begin with the expression:\n\n$$\\text{RHS} = 2 \\csc^2 \\phi$$\n\nRecall the fundamental reciprocal trigonometric identity for cosecant:\n\n$$\\csc \\phi = \\frac{1}{\\sin \\phi}$$\n\nSubstituting this definition into our expression gives:\n\n$$2 \\csc^2 \\phi = \\frac{2}{\\sin^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies \\htmlData{anchor=sin2-term}{\\color{@sky}{\\sin^2 \\phi}} = \\htmlData{anchor=pythag-sub}{\\color{@sky}{1 - \\cos^2 \\phi}}$$"},{"latex":"$$$\\frac{2}{\\htmlData{anchor=denom-sin}{\\color{@sky}{\\sin^2 \\phi}}} = \\frac{2}{\\htmlData{anchor=denom-cos}{\\color{@sky}{1 - \\cos^2 \\phi}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin2-term"},{"color":"sky","style":"text-color","anchor":"pythag-sub"}],"connections":[{"to":"denom-sin","from":"sin2-term","color":"sky","label":"substitute","style":"solid"},{"to":"denom-cos","from":"pythag-sub","color":"sky","label":"denominator","style":"solid"}]},"order":1,"title":"Use the Pythagorean identity","description":"We can rewrite the denominator in terms of cosine using the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$."},{"type":"text","order":2,"title":"Factorise the denominator","content":"Recognise that the denominator $1 - \\cos^2 \\phi$ is a difference of two squares. We can factorise it as follows:\n\n$$1 - \\cos^2 \\phi = (1 - \\cos \\phi)(1 + \\cos \\phi)$$\n\nSubstituting this back into the denominator yields:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = \\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":3,"title":"Decompose the fraction","content":"Now we split the single fraction into a sum of two separate terms. \n\nLet's rewrite the numerator $2$ as $1 + 1$. We can strategically add and subtract $\\cos \\phi$ in the numerator:\n\n$$2 = (1 + \\cos \\phi) + (1 - \\cos \\phi)$$\n\nNow rewrite the fraction with this split numerator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":4,"title":"Simplify to match LHS","content":"Separate the fraction into two parts:\n\n$$\\frac{1 + \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)} + \\frac{1 - \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nCancel the common terms in the numerator and denominator of each fraction:\n\n$$\\frac{\\cancel{1 + \\cos \\phi}}{(1 - \\cos \\phi)\\cancel{(1 + \\cos \\phi)}} + \\frac{\\cancel{1 - \\cos \\phi}}{\\cancel{(1 - \\cos \\phi)}(1 + \\cos \\phi)} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$\n\nThis is exactly the Left-Hand Side (LHS) of our equation, completing the proof:\n\n$$\\text{LHS} = \\text{RHS}$$"}]}],"generatedAt":"2026-06-09T04:39:08.710Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"f7b20b36-dae6-4182-a8ba-874cd43d14e6","label":"b","content":"The response \$R\$ of a specific sensor is modeled by the expression \$R = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}\$. Find the set of values of \$k\$ for which the equation \$R = k\$ has real solutions for \$\\phi\$. Fully justify your answer.","markscheme":"- Recognizes that for real solutions, $0 < \\sin^2 \\phi \\le 1$ since $\\cos \\phi \\neq \\pm 1$ M1\n- Deduces that $\\csc^2 \\phi = \\frac{1}{\\sin^2 \\phi} \\ge 1$, hence $k = 2 \\csc^2 \\phi \\ge 2$ A1","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"trigonometric_range","label":"Trigonometric Range Approach","steps":[{"type":"text","order":0,"title":"Simplify the expression","content":"To find the set of values for $k$ where $R = k$ has real solutions, we first simplify the expression for $R$ by combining the algebraic fractions over a common denominator: \n\n $$R = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$ \n\n Using a common denominator: \n\n $$\\begin{aligned} R &= \\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} \\\\ &= \\frac{2}{1 - \\cos^2 \\phi} \\end{aligned}$$ \n\n Using the fundamental trigonometric identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$, we substitute $1 - \\cos^2 \\phi = \\sin^2 \\phi$ to get: \n\n $$R = \\frac{2}{\\sin^2 \\phi}$$"},{"type":"text","order":1,"title":"Analyze the range","content":"For the expression $R = \\frac{2}{\\sin^2 \\phi}$ to have real solutions for $\\phi$, we need to consider the constraints on $\\sin^2 \\phi$. \n\n For any real angle $\\phi$: \n\n $$0 \\le \\sin^2 \\phi \\le 1$$ \n\n However, in the original expression for $R$, the denominators $1 - \\cos \\phi$ and $1 + \\cos \\phi$ cannot be zero. This means $\\cos \\phi \\neq \\pm 1$, which consequently means $\\sin \\phi \\neq 0$. Therefore, $\\sin^2 \\phi$ cannot equal $0$, which gives us the strict inequality: \n\n $$0 < \\sin^2 \\phi \\le 1$$"},{"type":"text","order":2,"title":"Determine values of k","content":"Now we set $R = k$ and solve for the range of $k$: \n\n $$k = \\frac{2}{\\sin^2 \\phi}$$ \n\n Applying the inequality $0 < \\sin^2 \\phi \\le 1$ by taking the reciprocal of each part: \n\n $$\\begin{aligned} \\frac{1}{\\sin^2 \\phi} &\\ge 1 \\\\ \\frac{2}{\\sin^2 \\phi} &\\ge 2 \\end{aligned}$$ \n\n Thus, we have: \n\n $$k \\ge 2$$ \n\n So, the set of values of $k$ for which the equation has real solutions is $k \\ge 2$."}]},{"id":"solving_for_cosine","label":"Solving for Cosine","steps":[{"type":"text","order":0,"title":"Equate R to k","content":"We start with the given expression for the sensor's response $R$: \n\n $$R = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$ \n\n To find the set of values of $k$ for which the equation $R = k$ has real solutions for $\\phi$, we set the expression equal to $k$: \n\n $$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = k$$"},{"type":"text","order":1,"title":"Combine the fractions","content":"To simplify the left side, we find a common denominator, which is $(1 - \\cos \\phi)(1 + \\cos \\phi)$: \n\n $$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = k$$ \n\n Simplifying the numerator gives: \n\n $$\\frac{2}{1 - \\cos^2 \\phi} = k$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2}{\\htmlData{anchor=denom}{\\color{@sky}{1 - \\cos^2 \\phi}}} = \\htmlData{anchor=k-val}{\\color{@amber}{k}}$"},{"latex":"$$\\htmlData{anchor=rearranged}{\\color{@sky}{1 - \\cos^2 \\phi}} = \\frac{2}{\\htmlData{anchor=k-denom}{\\color{@amber}{k}}}$"},{"latex":"$$\\htmlData{anchor=cos-sq}{\\color{@purple}{\\cos^2 \\phi}} = 1 - \\frac{2}{\\color{@amber}{k}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"denom"},{"color":"amber","style":"text-color","anchor":"k-val"},{"color":"purple","style":"text-color","anchor":"cos-sq"}],"connections":[{"to":"rearranged","from":"denom","color":"sky","label":null,"style":"solid"},{"to":"k-denom","from":"k-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Rearrange for cos²φ","description":"We rearrange the equation to express $\\cos^2 \\phi$ in terms of $k$."},{"type":"text","order":3,"title":"Apply domain restrictions","content":"For a real solution $\\phi$ to exist, we must have constraints on $\\cos \\phi$: \n\n 1. Since $\\cos \\phi$ is a real cosine value, we must have $-1 \\le \\cos \\phi \\le 1$, which implies: \n\n $$0 \\le \\cos^2 \\phi \\le 1$$ \n\n 2. However, the original equation has denominators of $1 - \\cos \\phi$ and $1 + \\cos \\phi$. These cannot be zero, which means $\\cos \\phi \\neq \\pm 1$, and thus $\\cos^2 \\phi \\neq 1$. \n\n Combining these two conditions gives the strict inequality: \n\n $$0 \\le \\cos^2 \\phi < 1$$"},{"type":"text","order":4,"title":"Solve the inequality","content":"Now substitute $\\cos^2 \\phi = 1 - \\frac{2}{k}$ into the inequality $0 \\le \\cos^2 \\phi < 1$: \n\n $$0 \\le 1 - \\frac{2}{k} < 1$$ \n\n Let's solve these two boundaries separately. \n\n First, consider the left-hand inequality: \n\n $$1 - \\frac{2}{k} \\ge 0 \\implies 1 \\ge \\frac{2}{k}$$ \n\n Second, consider the right-hand inequality: \n\n $$1 - \\frac{2}{k} < 1 \\implies -\\frac{2}{k} < 0 \\implies \\frac{2}{k} > 0$$ \n\n From $\\frac{2}{k} > 0$, we deduce that $k$ must be positive ($k > 0$). \n\n Since $k$ is positive, we can safely multiply both sides of the inequality $1 \\ge \\frac{2}{k}$ by $k$ without reversing the sign: \n\n $$k \\ge 2$$"},{"type":"text","order":5,"title":"State final answer","content":"By satisfying both conditions, we find that real solutions exist when: \n\n $$k \\ge 2$$"}]},{"id":"cosecant_range","label":"Cosecant Function Analysis","steps":[{"type":"text","order":0,"title":"Combine the fractions","content":"To simplify the expression for $R$, we first find a common denominator and add the two algebraic fractions: \n\n $$\\begin{aligned} R &= \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} \\\\ &= \\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} \\end{aligned}$$ \n\n Simplifying the numerator gives $2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R = \\frac{2}{\\htmlData{anchor=denom}{\\color{@sky}{1 - \\cos^2 \\phi}}}$"},{"latex":"$$\\text{Since } \\htmlData{anchor=identity}{\\color{@sky}{\\sin^2 \\phi}} + \\cos^2 \\phi = 1$"},{"latex":"$$R = \\frac{2}{\\htmlData{anchor=subbed}{\\color{@sky}{\\sin^2 \\phi}}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"denom"},{"color":"sky","style":"box","anchor":"identity"},{"color":"sky","style":"text-color","anchor":"subbed"}],"connections":[{"to":"identity","from":"denom","color":"sky","label":"identity","style":"dashed"},{"to":"subbed","from":"identity","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Use Pythagorean identity","description":"We recognize the denominator is a difference of squares: $(1 - \\cos \\phi)(1 + \\cos \\phi) = 1 - \\cos^2 \\phi$. We can then use the standard Pythagorean identity to replace it with $\\sin^2 \\phi$."},{"type":"text","order":2,"title":"Write as cosecant function","content":"Using the reciprocal trigonometric identity $\\csc \\phi = \\frac{1}{\\sin \\phi}$, we can rewrite the expression in terms of cosecant: \n\n $$R = 2 \\left(\\frac{1}{\\sin^2 \\phi}\\right) = 2\\csc^2 \\phi$$"},{"type":"text","order":3,"title":"Analyze range and solve","content":"For the equation $R = k$ to have real solutions for $\\phi$, $k$ must lie within the range of possible values for $R$. \n\n 1. First, note that for the expression to be defined, the denominator cannot be zero, which means $\\cos \\phi \\neq \\pm 1$ and thus $\\sin \\phi \\neq 0$. \n\n 2. Since $-1 \\le \\sin \\phi \\le 1$ and $\\sin \\phi \\neq 0$, we have: \n\n $$0 < \\sin^2 \\phi \\le 1$$ \n\n 3. Taking the reciprocal reverses the inequality: \n\n $$\\csc^2 \\phi = \\frac{1}{\\sin^2 \\phi} \\ge 1$$ \n\n 4. Multiplying by $2$: \n\n $$R = 2\\csc^2 \\phi \\ge 2$$ \n\n Therefore, the set of values of $k$ for which real solutions exist is $k \\ge 2$."}]}],"generatedAt":"2026-06-09T04:41:05.900Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"c3249fa6-7bda-4faf-a1e5-cee366b6c7de","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} = 8 \\]\nfind the exact value of \$\\cot \\phi\$.","markscheme":"- Uses the identity from part (a) to set $2 \\csc^2 \\phi = 8$, leading to $\\csc^2 \\phi = 4$ M1\n- Substitutes into the identity $1 + \\cot^2 \\phi = \\csc^2 \\phi$ to find $\\cot^2 \\phi = 3$ or $\\cot \\phi = \\pm\\sqrt{3}$ M1\n- States that $\\cot \\phi$ is positive in the third quadrant ($180^\\circ < \\phi < 270^\\circ$) and concludes $\\cot \\phi = \\sqrt{3}$ 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 equation","content":"We begin by simplifying the left-hand side of the given equation:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8$$\n\nTo add these fractions, we find a common denominator, which is $(1 - \\cos \\phi)(1 + \\cos \\phi) = 1 - \\cos^2 \\phi$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 8$$\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 8$$","highlights":[{"text":"\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the fundamental identity: } \\htmlData{anchor=trig-identity}{\\color{@sky}{1 - \\cos^2 \\phi = \\sin^2 \\phi}}$"},{"latex":"$$\\frac{2}{\\htmlData{anchor=denom-sub}{\\color{@sky}{\\sin^2 \\phi}}} = 8$"},{"latex":"$$2 \\htmlData{anchor=cosec-sub}{\\color{@amber}{\\csc^2 \\phi}} = 8 \\implies \\htmlData{anchor=cosec-val}{\\color{@amber}{\\csc^2 \\phi}} = 4$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"trig-identity"},{"color":"amber","style":"box","anchor":"cosec-val"}],"connections":[{"to":"denom-sub","from":"trig-identity","color":"sky","label":"substitute","style":"solid"},{"to":"cosec-val","from":"cosec-sub","color":"amber","label":"divide by 2","style":"solid"}]},"order":1,"title":"Express in cosecant","description":"Recall that $\\csc \\phi = \\frac{1}{\\sin \\phi}$, so we can write the equation in terms of $\\csc^2 \\phi$."},{"type":"text","order":2,"title":"Use Pythagorean identity","content":"Now we use the standard Pythagorean identity relating cosecant and cotangent:\n\n$$1 + \\cot^2 \\phi = \\csc^2 \\phi$$\n\nSubstituting our value of $\\csc^2 \\phi = 4$ into this identity gives:\n\n$$1 + \\cot^2 \\phi = 4$$\n\n$$\\cot^2 \\phi = 3$$\n\nTaking the square root of both sides, we get:\n\n$$\\cot \\phi = \\pm\\sqrt{3}$$"},{"type":"text","order":3,"title":"Determine the quadrant","content":"We are given that $\\phi$ lies in the third quadrant, where $180^\\circ < \\phi < 270^\\circ$.\n\nIn the third quadrant:\n* $\\sin \\phi < 0$\n* $\\cos \\phi < 0$\n\nSince $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$, dividing a negative number by a negative number results in a positive value. Thus, $\\cot \\phi$ must be positive in the third quadrant.\n\nTherefore, we choose the positive root:\n\n$$\\cot \\phi = \\sqrt{3}$$","highlights":[{"text":"third quadrant (reflex angle between 180° and 270°)","location":"specification"}]}]},{"id":"sine_cosine_ratio","label":"Solving for Sine and Cosine","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"To solve the equation \n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8$$\n\nwe first combine the fractions on the left-hand side by finding a common denominator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 8$$"},{"type":"text","order":1,"title":"Solve for sine","content":"Simplify the numerator and the denominator on the left side of the equation:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 8$$\n\nUsing the fundamental trigonometric identity $\\sin^2 \\phi = 1 - \\cos^2 \\phi$, we can substitute this into the denominator:\n\n$$\\frac{2}{\\sin^2 \\phi} = 8$$\n\nNow, solve for $\\sin^2 \\phi$:\n\n$$\\sin^2 \\phi = \\frac{2}{8} = \\frac{1}{4}$$\n\nTaking the square root of both sides gives:\n\n$$\\sin \\phi = \\pm\\frac{1}{2}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"We are given that $\\phi$ lies in the third quadrant ($180^\\circ < \\phi < 270^\\circ$). In this quadrant, both sine and cosine are negative."},{"latex":"$$\\htmlData{anchor=sin-val}{\\sin \\phi} = \\color{@red}{-\\frac{1}{2}}$"},{"latex":"$$\\cos^2 \\phi = 1 - \\sin^2 \\phi = 1 - \\left(-\\frac{1}{2}\\right)^2 = \\frac{3}{4}$"},{"latex":"$$\\htmlData{anchor=cos-val}{\\cos \\phi} = \\color{@blue}{-\\frac{\\sqrt{3}}{2}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"sin-val"},{"color":"indigo","style":"text-color","anchor":"cos-val"}],"connections":[]},"order":2,"title":"Determine signs in Q3","description":"We identify the signs of $\\sin\\phi$ and $\\cos\\phi$ based on the quadrant constraints and solve for their individual exact values."},{"type":"text","order":3,"title":"Calculate cotangent","content":"Now we use the quotient identity for $\\cot \\phi$:\n\n$$\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$$\n\nSubstitute our values for $\\sin \\phi$ and $\\cos \\phi$:\n\n$$\\cot \\phi = \\frac{-\\frac{\\sqrt{3}}{2}}{-\\frac{1}{2}} = \\sqrt{3}$$\n\nSince both $\\sin \\phi$ and $\\cos \\phi$ are negative, their ratio $\\cot \\phi$ is positive, which is correct for an angle in the third quadrant."}]}],"generatedAt":"2026-06-11T03:13:18.838Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4ebf97d0-72be-4d49-aa7a-6201d26b7f1b:lhs_to_rhs","label":"(a) Starting from LHS","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we start with the Left-Hand Side (LHS) of the equation:\n\n$$\\text{LHS} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$"},{"type":"text","order":1,"title":"Combine the fractions","content":"To combine the two algebraic fractions, we find a common denominator by multiplying the two denominators: $(1 - \\cos \\phi)(1 + \\cos \\phi)$.\n\nWe rewrite each fraction with this common denominator:\n\n$$\\frac{1 \\cdot (1 + \\cos \\phi) + 1 \\cdot (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nThis step earns the first method mark (**M1**)."},{"type":"text","order":2,"title":"Simplify numerator and denominator","content":"Now, let's simplify both the numerator and the denominator:\n\n1. **Numerator:** Combine the terms in the numerator:\n $$(1 + \\cos \\phi) + (1 - \\cos \\phi) = 2$$\n\n2. **Denominator:** Expand using the difference of two squares identity, $(a-b)(a+b) = a^2 - b^2$:\n $$(1 - \\cos \\phi)(1 + \\cos \\phi) = 1^2 - \\cos^2 \\phi = 1 - \\cos^2 \\phi$$\n\nSubstituting these back into our expression gives:\n\n$$\\frac{2}{1 - \\cos^2 \\phi}$$\n\nSimplifying to this form earns the second method mark (**M1**)."},{"type":"text","order":3,"title":"Apply trigonometric identities","content":"Recall the fundamental Pythagorean identity:\n\n$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies 1 - \\cos^2 \\phi = \\sin^2 \\phi$$\n\nSubstitute this into the denominator:\n\n$$\\frac{2}{\\sin^2 \\phi}$$\n\nNext, use the reciprocal trigonometric identity for cosecant, $\\csc \\phi = \\frac{1}{\\sin \\phi}$, which also means $\\csc^2 \\phi = \\frac{1}{\\sin^2 \\phi}$:\n\n$$\\frac{2}{\\sin^2 \\phi} = 2 \\left(\\frac{1}{\\sin^2 \\phi}\\right) = 2 \\csc^2 \\phi$$\n\nThis is equal to the Right-Hand Side (RHS). Showing this substitution clearly to reach the final expression completes the proof and earns the final accuracy mark (**A1**)."}]},{"id":"4ebf97d0-72be-4d49-aa7a-6201d26b7f1b:rhs_to_lhs","label":"(a) Starting from RHS","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity by starting from the Right-Hand Side (RHS), we begin with the expression:\n\n$$\\text{RHS} = 2 \\csc^2 \\phi$$\n\nRecall the fundamental reciprocal trigonometric identity for cosecant:\n\n$$\\csc \\phi = \\frac{1}{\\sin \\phi}$$\n\nSubstituting this definition into our expression gives:\n\n$$2 \\csc^2 \\phi = \\frac{2}{\\sin^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\sin^2 \\phi + \\cos^2 \\phi = 1 \\implies \\htmlData{anchor=sin2-term}{\\color{@sky}{\\sin^2 \\phi}} = \\htmlData{anchor=pythag-sub}{\\color{@sky}{1 - \\cos^2 \\phi}}$$"},{"latex":"$$$\\frac{2}{\\htmlData{anchor=denom-sin}{\\color{@sky}{\\sin^2 \\phi}}} = \\frac{2}{\\htmlData{anchor=denom-cos}{\\color{@sky}{1 - \\cos^2 \\phi}}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin2-term"},{"color":"sky","style":"text-color","anchor":"pythag-sub"}],"connections":[{"to":"denom-sin","from":"sin2-term","color":"sky","label":"substitute","style":"solid"},{"to":"denom-cos","from":"pythag-sub","color":"sky","label":"denominator","style":"solid"}]},"order":1,"title":"Use the Pythagorean identity","description":"We can rewrite the denominator in terms of cosine using the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$."},{"type":"text","order":2,"title":"Factorise the denominator","content":"Recognise that the denominator $1 - \\cos^2 \\phi$ is a difference of two squares. We can factorise it as follows:\n\n$$1 - \\cos^2 \\phi = (1 - \\cos \\phi)(1 + \\cos \\phi)$$\n\nSubstituting this back into the denominator yields:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = \\frac{2}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":3,"title":"Decompose the fraction","content":"Now we split the single fraction into a sum of two separate terms. \n\nLet's rewrite the numerator $2$ as $1 + 1$. We can strategically add and subtract $\\cos \\phi$ in the numerator:\n\n$$2 = (1 + \\cos \\phi) + (1 - \\cos \\phi)$$\n\nNow rewrite the fraction with this split numerator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$"},{"type":"text","order":4,"title":"Simplify to match LHS","content":"Separate the fraction into two parts:\n\n$$\\frac{1 + \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)} + \\frac{1 - \\cos \\phi}{(1 - \\cos \\phi)(1 + \\cos \\phi)}$$\n\nCancel the common terms in the numerator and denominator of each fraction:\n\n$$\\frac{\\cancel{1 + \\cos \\phi}}{(1 - \\cos \\phi)\\cancel{(1 + \\cos \\phi)}} + \\frac{\\cancel{1 - \\cos \\phi}}{\\cancel{(1 - \\cos \\phi)}(1 + \\cos \\phi)} = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$\n\nThis is exactly the Left-Hand Side (LHS) of our equation, completing the proof:\n\n$$\\text{LHS} = \\text{RHS}$$"}]},{"id":"f7b20b36-dae6-4182-a8ba-874cd43d14e6:trigonometric_range","label":"(b) Trigonometric Range Approach","steps":[{"type":"text","order":0,"title":"Simplify the expression","content":"To find the set of values for $k$ where $R = k$ has real solutions, we first simplify the expression for $R$ by combining the algebraic fractions over a common denominator: \n\n $$R = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$ \n\n Using a common denominator: \n\n $$\\begin{aligned} R &= \\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} \\\\ &= \\frac{2}{1 - \\cos^2 \\phi} \\end{aligned}$$ \n\n Using the fundamental trigonometric identity $\\sin^2 \\phi + \\cos^2 \\phi = 1$, we substitute $1 - \\cos^2 \\phi = \\sin^2 \\phi$ to get: \n\n $$R = \\frac{2}{\\sin^2 \\phi}$$"},{"type":"text","order":1,"title":"Analyze the range","content":"For the expression $R = \\frac{2}{\\sin^2 \\phi}$ to have real solutions for $\\phi$, we need to consider the constraints on $\\sin^2 \\phi$. \n\n For any real angle $\\phi$: \n\n $$0 \\le \\sin^2 \\phi \\le 1$$ \n\n However, in the original expression for $R$, the denominators $1 - \\cos \\phi$ and $1 + \\cos \\phi$ cannot be zero. This means $\\cos \\phi \\neq \\pm 1$, which consequently means $\\sin \\phi \\neq 0$. Therefore, $\\sin^2 \\phi$ cannot equal $0$, which gives us the strict inequality: \n\n $$0 < \\sin^2 \\phi \\le 1$$"},{"type":"text","order":2,"title":"Determine values of k","content":"Now we set $R = k$ and solve for the range of $k$: \n\n $$k = \\frac{2}{\\sin^2 \\phi}$$ \n\n Applying the inequality $0 < \\sin^2 \\phi \\le 1$ by taking the reciprocal of each part: \n\n $$\\begin{aligned} \\frac{1}{\\sin^2 \\phi} &\\ge 1 \\\\ \\frac{2}{\\sin^2 \\phi} &\\ge 2 \\end{aligned}$$ \n\n Thus, we have: \n\n $$k \\ge 2$$ \n\n So, the set of values of $k$ for which the equation has real solutions is $k \\ge 2$."}]},{"id":"f7b20b36-dae6-4182-a8ba-874cd43d14e6:solving_for_cosine","label":"(b) Solving for Cosine","steps":[{"type":"text","order":0,"title":"Equate R to k","content":"We start with the given expression for the sensor's response $R$: \n\n $$R = \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi}$$ \n\n To find the set of values of $k$ for which the equation $R = k$ has real solutions for $\\phi$, we set the expression equal to $k$: \n\n $$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = k$$"},{"type":"text","order":1,"title":"Combine the fractions","content":"To simplify the left side, we find a common denominator, which is $(1 - \\cos \\phi)(1 + \\cos \\phi)$: \n\n $$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = k$$ \n\n Simplifying the numerator gives: \n\n $$\\frac{2}{1 - \\cos^2 \\phi} = k$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{2}{\\htmlData{anchor=denom}{\\color{@sky}{1 - \\cos^2 \\phi}}} = \\htmlData{anchor=k-val}{\\color{@amber}{k}}$"},{"latex":"$$\\htmlData{anchor=rearranged}{\\color{@sky}{1 - \\cos^2 \\phi}} = \\frac{2}{\\htmlData{anchor=k-denom}{\\color{@amber}{k}}}$"},{"latex":"$$\\htmlData{anchor=cos-sq}{\\color{@purple}{\\cos^2 \\phi}} = 1 - \\frac{2}{\\color{@amber}{k}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"denom"},{"color":"amber","style":"text-color","anchor":"k-val"},{"color":"purple","style":"text-color","anchor":"cos-sq"}],"connections":[{"to":"rearranged","from":"denom","color":"sky","label":null,"style":"solid"},{"to":"k-denom","from":"k-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Rearrange for cos²φ","description":"We rearrange the equation to express $\\cos^2 \\phi$ in terms of $k$."},{"type":"text","order":3,"title":"Apply domain restrictions","content":"For a real solution $\\phi$ to exist, we must have constraints on $\\cos \\phi$: \n\n 1. Since $\\cos \\phi$ is a real cosine value, we must have $-1 \\le \\cos \\phi \\le 1$, which implies: \n\n $$0 \\le \\cos^2 \\phi \\le 1$$ \n\n 2. However, the original equation has denominators of $1 - \\cos \\phi$ and $1 + \\cos \\phi$. These cannot be zero, which means $\\cos \\phi \\neq \\pm 1$, and thus $\\cos^2 \\phi \\neq 1$. \n\n Combining these two conditions gives the strict inequality: \n\n $$0 \\le \\cos^2 \\phi < 1$$"},{"type":"text","order":4,"title":"Solve the inequality","content":"Now substitute $\\cos^2 \\phi = 1 - \\frac{2}{k}$ into the inequality $0 \\le \\cos^2 \\phi < 1$: \n\n $$0 \\le 1 - \\frac{2}{k} < 1$$ \n\n Let's solve these two boundaries separately. \n\n First, consider the left-hand inequality: \n\n $$1 - \\frac{2}{k} \\ge 0 \\implies 1 \\ge \\frac{2}{k}$$ \n\n Second, consider the right-hand inequality: \n\n $$1 - \\frac{2}{k} < 1 \\implies -\\frac{2}{k} < 0 \\implies \\frac{2}{k} > 0$$ \n\n From $\\frac{2}{k} > 0$, we deduce that $k$ must be positive ($k > 0$). \n\n Since $k$ is positive, we can safely multiply both sides of the inequality $1 \\ge \\frac{2}{k}$ by $k$ without reversing the sign: \n\n $$k \\ge 2$$"},{"type":"text","order":5,"title":"State final answer","content":"By satisfying both conditions, we find that real solutions exist when: \n\n $$k \\ge 2$$"}]},{"id":"f7b20b36-dae6-4182-a8ba-874cd43d14e6:cosecant_range","label":"(b) Cosecant Function Analysis","steps":[{"type":"text","order":0,"title":"Combine the fractions","content":"To simplify the expression for $R$, we first find a common denominator and add the two algebraic fractions: \n\n $$\\begin{aligned} R &= \\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} \\\\ &= \\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} \\end{aligned}$$ \n\n Simplifying the numerator gives $2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R = \\frac{2}{\\htmlData{anchor=denom}{\\color{@sky}{1 - \\cos^2 \\phi}}}$"},{"latex":"$$\\text{Since } \\htmlData{anchor=identity}{\\color{@sky}{\\sin^2 \\phi}} + \\cos^2 \\phi = 1$"},{"latex":"$$R = \\frac{2}{\\htmlData{anchor=subbed}{\\color{@sky}{\\sin^2 \\phi}}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"denom"},{"color":"sky","style":"box","anchor":"identity"},{"color":"sky","style":"text-color","anchor":"subbed"}],"connections":[{"to":"identity","from":"denom","color":"sky","label":"identity","style":"dashed"},{"to":"subbed","from":"identity","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Use Pythagorean identity","description":"We recognize the denominator is a difference of squares: $(1 - \\cos \\phi)(1 + \\cos \\phi) = 1 - \\cos^2 \\phi$. We can then use the standard Pythagorean identity to replace it with $\\sin^2 \\phi$."},{"type":"text","order":2,"title":"Write as cosecant function","content":"Using the reciprocal trigonometric identity $\\csc \\phi = \\frac{1}{\\sin \\phi}$, we can rewrite the expression in terms of cosecant: \n\n $$R = 2 \\left(\\frac{1}{\\sin^2 \\phi}\\right) = 2\\csc^2 \\phi$$"},{"type":"text","order":3,"title":"Analyze range and solve","content":"For the equation $R = k$ to have real solutions for $\\phi$, $k$ must lie within the range of possible values for $R$. \n\n 1. First, note that for the expression to be defined, the denominator cannot be zero, which means $\\cos \\phi \\neq \\pm 1$ and thus $\\sin \\phi \\neq 0$. \n\n 2. Since $-1 \\le \\sin \\phi \\le 1$ and $\\sin \\phi \\neq 0$, we have: \n\n $$0 < \\sin^2 \\phi \\le 1$$ \n\n 3. Taking the reciprocal reverses the inequality: \n\n $$\\csc^2 \\phi = \\frac{1}{\\sin^2 \\phi} \\ge 1$$ \n\n 4. Multiplying by $2$: \n\n $$R = 2\\csc^2 \\phi \\ge 2$$ \n\n Therefore, the set of values of $k$ for which real solutions exist is $k \\ge 2$."}]},{"id":"c3249fa6-7bda-4faf-a1e5-cee366b6c7de:pythagorean_identity","label":"(c) Using Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"We begin by simplifying the left-hand side of the given equation:\n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8$$\n\nTo add these fractions, we find a common denominator, which is $(1 - \\cos \\phi)(1 + \\cos \\phi) = 1 - \\cos^2 \\phi$:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 8$$\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 8$$","highlights":[{"text":"\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the fundamental identity: } \\htmlData{anchor=trig-identity}{\\color{@sky}{1 - \\cos^2 \\phi = \\sin^2 \\phi}}$"},{"latex":"$$\\frac{2}{\\htmlData{anchor=denom-sub}{\\color{@sky}{\\sin^2 \\phi}}} = 8$"},{"latex":"$$2 \\htmlData{anchor=cosec-sub}{\\color{@amber}{\\csc^2 \\phi}} = 8 \\implies \\htmlData{anchor=cosec-val}{\\color{@amber}{\\csc^2 \\phi}} = 4$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"trig-identity"},{"color":"amber","style":"box","anchor":"cosec-val"}],"connections":[{"to":"denom-sub","from":"trig-identity","color":"sky","label":"substitute","style":"solid"},{"to":"cosec-val","from":"cosec-sub","color":"amber","label":"divide by 2","style":"solid"}]},"order":1,"title":"Express in cosecant","description":"Recall that $\\csc \\phi = \\frac{1}{\\sin \\phi}$, so we can write the equation in terms of $\\csc^2 \\phi$."},{"type":"text","order":2,"title":"Use Pythagorean identity","content":"Now we use the standard Pythagorean identity relating cosecant and cotangent:\n\n$$1 + \\cot^2 \\phi = \\csc^2 \\phi$$\n\nSubstituting our value of $\\csc^2 \\phi = 4$ into this identity gives:\n\n$$1 + \\cot^2 \\phi = 4$$\n\n$$\\cot^2 \\phi = 3$$\n\nTaking the square root of both sides, we get:\n\n$$\\cot \\phi = \\pm\\sqrt{3}$$"},{"type":"text","order":3,"title":"Determine the quadrant","content":"We are given that $\\phi$ lies in the third quadrant, where $180^\\circ < \\phi < 270^\\circ$.\n\nIn the third quadrant:\n* $\\sin \\phi < 0$\n* $\\cos \\phi < 0$\n\nSince $\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$, dividing a negative number by a negative number results in a positive value. Thus, $\\cot \\phi$ must be positive in the third quadrant.\n\nTherefore, we choose the positive root:\n\n$$\\cot \\phi = \\sqrt{3}$$","highlights":[{"text":"third quadrant (reflex angle between 180° and 270°)","location":"specification"}]}]},{"id":"c3249fa6-7bda-4faf-a1e5-cee366b6c7de:sine_cosine_ratio","label":"(c) Solving for Sine and Cosine","steps":[{"type":"text","order":0,"title":"Simplify the equation","content":"To solve the equation \n\n$$\\frac{1}{1 - \\cos \\phi} + \\frac{1}{1 + \\cos \\phi} = 8$$\n\nwe first combine the fractions on the left-hand side by finding a common denominator:\n\n$$\\frac{(1 + \\cos \\phi) + (1 - \\cos \\phi)}{(1 - \\cos \\phi)(1 + \\cos \\phi)} = 8$$"},{"type":"text","order":1,"title":"Solve for sine","content":"Simplify the numerator and the denominator on the left side of the equation:\n\n$$\\frac{2}{1 - \\cos^2 \\phi} = 8$$\n\nUsing the fundamental trigonometric identity $\\sin^2 \\phi = 1 - \\cos^2 \\phi$, we can substitute this into the denominator:\n\n$$\\frac{2}{\\sin^2 \\phi} = 8$$\n\nNow, solve for $\\sin^2 \\phi$:\n\n$$\\sin^2 \\phi = \\frac{2}{8} = \\frac{1}{4}$$\n\nTaking the square root of both sides gives:\n\n$$\\sin \\phi = \\pm\\frac{1}{2}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"We are given that $\\phi$ lies in the third quadrant ($180^\\circ < \\phi < 270^\\circ$). In this quadrant, both sine and cosine are negative."},{"latex":"$$\\htmlData{anchor=sin-val}{\\sin \\phi} = \\color{@red}{-\\frac{1}{2}}$"},{"latex":"$$\\cos^2 \\phi = 1 - \\sin^2 \\phi = 1 - \\left(-\\frac{1}{2}\\right)^2 = \\frac{3}{4}$"},{"latex":"$$\\htmlData{anchor=cos-val}{\\cos \\phi} = \\color{@blue}{-\\frac{\\sqrt{3}}{2}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"sin-val"},{"color":"indigo","style":"text-color","anchor":"cos-val"}],"connections":[]},"order":2,"title":"Determine signs in Q3","description":"We identify the signs of $\\sin\\phi$ and $\\cos\\phi$ based on the quadrant constraints and solve for their individual exact values."},{"type":"text","order":3,"title":"Calculate cotangent","content":"Now we use the quotient identity for $\\cot \\phi$:\n\n$$\\cot \\phi = \\frac{\\cos \\phi}{\\sin \\phi}$$\n\nSubstitute our values for $\\sin \\phi$ and $\\cos \\phi$:\n\n$$\\cot \\phi = \\frac{-\\frac{\\sqrt{3}}{2}}{-\\frac{1}{2}} = \\sqrt{3}$$\n\nSince both $\\sin \\phi$ and $\\cos \\phi$ are negative, their ratio $\\cot \\phi$ is positive, which is correct for an angle in the third quadrant."}]}],"generatedAt":"2026-06-11T03:13:18.888Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"517d3ea0-a5c6-4af9-86f3-527f74dc0e63","revisionId":"11abd873-ef75-47bd-85c1-9e371329f71d","questionNumber":null,"questionType":"LA","specification":"Solve, for $0° < x < 360°$, the equation\n\n$\\operatorname{cosec}^2 x - 2\\cot x = 4$","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"e3ac4d4e-7a61-423d-842f-71adadc0477e","label":null,"content":"Give your answers to one decimal place where appropriate.","markscheme":"- Uses $\\operatorname{cosec}^2 x \\equiv 1 + \\cot^2 x$ M1\n- $\\cot^2 x - 2\\cot x - 3 = 0$ A1\n- $(\\cot x - 3)(\\cot x + 1) = 0$ M1\n- $\\cot x = 3$ gives $\\tan x = \\frac{1}{3}$, so $x = 18.4°$ and $x = 198.4°$ A1\n- $\\cot x = -1$ gives $\\tan x = -1$ M1\n- $x = 135°$ and $x = 315°$ A1\n\nAll four values are required.","marks":6,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"51ccf6f4-67c2-4018-954e-431375c0100b","revisionId":"96740f66-ef11-436e-82cb-f7f156d7b05c","questionNumber":null,"questionType":"LA","specification":"Solve, for $0 \\leq \\theta < \\pi$, the equation\n\n$4\\cos\\theta = \\operatorname{cosec}\\theta$","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"66899eb7-21df-4dd1-8bbc-198e5071d4f7","label":null,"content":"Give your answers in terms of $\\pi$.","markscheme":"- Writes $\\operatorname{cosec}\\theta = \\frac{1}{\\sin\\theta}$ and multiplies through: $4\\sin\\theta\\cos\\theta = 1$ M1\n- Uses $2\\sin\\theta\\cos\\theta \\equiv \\sin 2\\theta$: $2\\sin 2\\theta = 1$ M1\n- $\\sin 2\\theta = \\frac{1}{2}$ A1\n- $2\\theta = \\frac{\\pi}{6}$ or $\\frac{5\\pi}{6}$ M1\n- $\\theta = \\frac{\\pi}{12}$ and $\\theta = \\frac{5\\pi}{12}$ A1\n\nThe interval for $2\\theta$ is $0 \\leq 2\\theta < 2\\pi$. Multiplying by $\\sin\\theta$ requires $\\sin\\theta \\neq 0$, which both solutions 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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 82

3.4 Trigonometry (A-level only) Questions

3.4 Trigonometry (A-level only)

Question 82

3.4 Trigonometry (A-level only) Questions