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A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Substitute the double angle identity $\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$ in the numerator.","code":"M"},{"id":"p2","marks":1,"description":"Correctly factorise the numerator as a difference of squares and cancel the denominator to reach the final expression.","code":"A","acceptable":["LHS = (cosθ - sinθ)(cosθ + sinθ) / (cosθ - sinθ) = cosθ + sinθ","Demonstrates factorisation and explicit cancellation of the (cosθ - sinθ) term."]}],"finalAnswers":["\\cos \\theta + \\sin \\theta"],"generatedAt":"2026-05-06T14:03:51.189Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"lhs_to_rhs","label":"LHS to RHS Expansion","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we start with the Left-Hand Side (LHS):\n\n$$\\text{LHS} = \\frac{\\cos 2\\theta}{\\cos \\theta - \\sin \\theta}$$\n\nOur goal is to manipulate the numerator so that we can simplify this fraction to match the Right-Hand Side (RHS)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Recall: } \\htmlData{anchor=double-angle-cos}{\\color{@sky}{\\cos 2\\theta}} \\equiv \\htmlData{anchor=diff-squares}{\\color{@sky}{\\cos^2 \\theta - \\sin^2 \\theta}}$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=sub-num}{\\color{@sky}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"double-angle-cos"},{"color":"sky","style":"underline","anchor":"sub-num"}],"connections":[{"to":"sub-num","from":"diff-squares","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute double-angle identity","description":"We use the double-angle identity for cosine from our formula sheet to rewrite the numerator. Substituting $$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$ earns the first method mark **(M1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using } a^2 - b^2 = (a - b)(a + b):$"},{"latex":"$$\\htmlData{anchor=unfactored}{\\color{@amber}{\\cos^2 \\theta - \\sin^2 \\theta}} = \\htmlData{anchor=factored}{\\color{@amber}{(\\cos \\theta - \\sin \\theta)(\\cos \\theta + \\sin \\theta)}}$"},{"latex":"$$\\text{LHS} = \\frac{\\color{@amber}{(\\cos \\theta - \\sin \\theta)(\\cos \\theta + \\sin \\theta)}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"unfactored"},{"color":"amber","style":"box","anchor":"factored"}],"connections":[{"to":"factored","from":"unfactored","color":"amber","label":"factorise","style":"solid"}]},"order":2,"title":"Factorise the numerator","description":"The numerator is now a difference of two squares. We can factorise it in the form $$(a-b)(a+b)$$, where $$a = \\cos \\theta$$ and $$b = \\sin \\theta$$."},{"type":"text","order":3,"title":"Simplify and reach RHS","content":"Now, we can cancel out the common factor of $(\\cos \\theta - \\sin \\theta)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\frac{\\cancel{(\\cos \\theta - \\sin \\theta)}(\\cos \\theta + \\sin \\theta)}{\\cancel{\\cos \\theta - \\sin \\theta}}$$\n\n$$\\text{LHS} = \\cos \\theta + \\sin \\theta = \\text{RHS}$$\n\nCompleting the factorisation and showing the cancellation to reach the final expression earns the final accuracy mark **(A1)**."}]},{"id":"rhs_to_lhs","label":"RHS to LHS Multiplication","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using the requested approach, we begin with the right-hand side (RHS) of the equation:\n\n$$\\text{RHS} = \\cos \\theta + \\sin \\theta$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\cos \\theta + \\sin \\theta$"},{"latex":"$$\\text{RHS} = \\frac{(\\cos \\theta + \\sin \\theta)\\htmlData{anchor=conjugate-num}{\\color{@sky}{(\\cos \\theta - \\sin \\theta)}}}{\\htmlData{anchor=conjugate-den}{\\color{@sky}{\\cos \\theta - \\sin \\theta}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"conjugate-num"},{"color":"sky","style":"text-color","anchor":"conjugate-den"}],"connections":[{"to":"conjugate-den","from":"conjugate-num","color":"sky","label":"conjugate","style":"solid"}]},"order":1,"title":"Multiply by the conjugate","description":"We multiply and divide the RHS expression by the conjugate term, which is $(\\cos \\theta - \\sin \\theta)$. This sets us up to create a difference of squares in the numerator."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=expanded-num}{\\color{@amber}{(\\cos \\theta + \\sin \\theta)(\\cos \\theta - \\sin \\theta)}}}{\\cos \\theta - \\sin \\theta}$"},{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=diff-sq}{\\color{@amber}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"expanded-num"},{"color":"amber","style":"text-color","anchor":"diff-sq"}],"connections":[{"to":"diff-sq","from":"expanded-num","color":"amber","label":"difference of squares","style":"solid"}]},"order":2,"title":"Expand the numerator","description":"We expand the numerator using the difference of two squares identity, $(a + b)(a - b) = a^2 - b^2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=cos-diff}{\\color{@purple}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"},{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=cos-double}{\\color{@purple}{\\cos 2\\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"cos-diff"},{"color":"purple","style":"text-color","anchor":"cos-double"}],"connections":[{"to":"cos-double","from":"cos-diff","color":"purple","label":"double-angle identity","style":"solid"}]},"order":3,"title":"Apply double-angle identity","description":"Now, we recall the double-angle identity for cosine: $$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$ We substitute this identity into the numerator."},{"type":"text","order":4,"title":"State the conclusion","content":"This directly matches the left-hand side (LHS) of our original equation:\n\n$$\\text{RHS} = \\frac{\\cos 2\\theta}{\\cos \\theta - \\sin \\theta} = \\text{LHS}$$\n\nThus, the identity has been proven."}]}],"generatedAt":"2026-06-09T01:58:31.975Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"0ad76e48-049b-4e33-8e3e-2667ec1b0213:lhs_to_rhs","label":"LHS to RHS Expansion","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we start with the Left-Hand Side (LHS):\n\n$$\\text{LHS} = \\frac{\\cos 2\\theta}{\\cos \\theta - \\sin \\theta}$$\n\nOur goal is to manipulate the numerator so that we can simplify this fraction to match the Right-Hand Side (RHS)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Recall: } \\htmlData{anchor=double-angle-cos}{\\color{@sky}{\\cos 2\\theta}} \\equiv \\htmlData{anchor=diff-squares}{\\color{@sky}{\\cos^2 \\theta - \\sin^2 \\theta}}$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=sub-num}{\\color{@sky}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"double-angle-cos"},{"color":"sky","style":"underline","anchor":"sub-num"}],"connections":[{"to":"sub-num","from":"diff-squares","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute double-angle identity","description":"We use the double-angle identity for cosine from our formula sheet to rewrite the numerator. Substituting $$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$ earns the first method mark **(M1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using } a^2 - b^2 = (a - b)(a + b):$"},{"latex":"$$\\htmlData{anchor=unfactored}{\\color{@amber}{\\cos^2 \\theta - \\sin^2 \\theta}} = \\htmlData{anchor=factored}{\\color{@amber}{(\\cos \\theta - \\sin \\theta)(\\cos \\theta + \\sin \\theta)}}$"},{"latex":"$$\\text{LHS} = \\frac{\\color{@amber}{(\\cos \\theta - \\sin \\theta)(\\cos \\theta + \\sin \\theta)}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"unfactored"},{"color":"amber","style":"box","anchor":"factored"}],"connections":[{"to":"factored","from":"unfactored","color":"amber","label":"factorise","style":"solid"}]},"order":2,"title":"Factorise the numerator","description":"The numerator is now a difference of two squares. We can factorise it in the form $$(a-b)(a+b)$$, where $$a = \\cos \\theta$$ and $$b = \\sin \\theta$$."},{"type":"text","order":3,"title":"Simplify and reach RHS","content":"Now, we can cancel out the common factor of $(\\cos \\theta - \\sin \\theta)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\frac{\\cancel{(\\cos \\theta - \\sin \\theta)}(\\cos \\theta + \\sin \\theta)}{\\cancel{\\cos \\theta - \\sin \\theta}}$$\n\n$$\\text{LHS} = \\cos \\theta + \\sin \\theta = \\text{RHS}$$\n\nCompleting the factorisation and showing the cancellation to reach the final expression earns the final accuracy mark **(A1)**."}]},{"id":"0ad76e48-049b-4e33-8e3e-2667ec1b0213:rhs_to_lhs","label":"RHS to LHS Multiplication","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using the requested approach, we begin with the right-hand side (RHS) of the equation:\n\n$$\\text{RHS} = \\cos \\theta + \\sin \\theta$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\cos \\theta + \\sin \\theta$"},{"latex":"$$\\text{RHS} = \\frac{(\\cos \\theta + \\sin \\theta)\\htmlData{anchor=conjugate-num}{\\color{@sky}{(\\cos \\theta - \\sin \\theta)}}}{\\htmlData{anchor=conjugate-den}{\\color{@sky}{\\cos \\theta - \\sin \\theta}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"conjugate-num"},{"color":"sky","style":"text-color","anchor":"conjugate-den"}],"connections":[{"to":"conjugate-den","from":"conjugate-num","color":"sky","label":"conjugate","style":"solid"}]},"order":1,"title":"Multiply by the conjugate","description":"We multiply and divide the RHS expression by the conjugate term, which is $(\\cos \\theta - \\sin \\theta)$. This sets us up to create a difference of squares in the numerator."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=expanded-num}{\\color{@amber}{(\\cos \\theta + \\sin \\theta)(\\cos \\theta - \\sin \\theta)}}}{\\cos \\theta - \\sin \\theta}$"},{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=diff-sq}{\\color{@amber}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"expanded-num"},{"color":"amber","style":"text-color","anchor":"diff-sq"}],"connections":[{"to":"diff-sq","from":"expanded-num","color":"amber","label":"difference of squares","style":"solid"}]},"order":2,"title":"Expand the numerator","description":"We expand the numerator using the difference of two squares identity, $(a + b)(a - b) = a^2 - b^2$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=cos-diff}{\\color{@purple}{\\cos^2 \\theta - \\sin^2 \\theta}}}{\\cos \\theta - \\sin \\theta}$"},{"latex":"$$\\text{RHS} = \\frac{\\htmlData{anchor=cos-double}{\\color{@purple}{\\cos 2\\theta}}}{\\cos \\theta - \\sin \\theta}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"cos-diff"},{"color":"purple","style":"text-color","anchor":"cos-double"}],"connections":[{"to":"cos-double","from":"cos-diff","color":"purple","label":"double-angle identity","style":"solid"}]},"order":3,"title":"Apply double-angle identity","description":"Now, we recall the double-angle identity for cosine: $$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$ We substitute this identity into the numerator."},{"type":"text","order":4,"title":"State the conclusion","content":"This directly matches the left-hand side (LHS) of our original equation:\n\n$$\\text{RHS} = \\frac{\\cos 2\\theta}{\\cos \\theta - \\sin \\theta} = \\text{LHS}$$\n\nThus, the identity has been proven."}]}],"generatedAt":"2026-06-09T01:59:08.716Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":65},"walkthroughVideo":null},{"id":"327076de-b61b-400a-bde2-937a30472d46","revisionId":"923d49fa-039b-4deb-a28d-e015a3ef854e","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2b72a823-4676-4a0e-933d-bdced993c272","label":"a","content":"Prove the identity:\n\n$$\\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x} \\equiv \\cos x + \\sin x$$","markscheme":"- Recognises $1 - 2\\sin^2 x = \\cos 2x$ and uses $\\cos 2x = \\cos^2 x - \\sin^2 x$ OR substitutes $1 = \\cos^2 x + \\sin^2 x$ M1\n- Simplifies numerator to $\\cos^2 x - \\sin^2 x$, factorises, and cancels to reach RHS A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Apply the identity $1 - 2\\sin^2 x = \\cos^2 x - \\sin^2 x$ or substitute $1 = \\sin^2 x + \\cos^2 x$.","code":"M"},{"id":"p2","marks":1,"description":"Factorise the resulting numerator as $(\\cos x - \\sin x)(\\cos x + \\sin x)$ and simplify the fraction.","code":"A","acceptable":["LHS = (cos^2 x - sin^2 x) / (cos x - sin x) = cos x + sin x"]}],"finalAnswers":["\\cos x + \\sin x"],"generatedAt":"2026-05-06T14:03:52.906Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"pythagorean_substitution","label":"Pythagorean Identity Substitution","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression:\n\n$$\\text{LHS} = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":1,"title":"Substitute the Pythagorean Identity","content":"Recall the fundamental Pythagorean identity:\n\n$$\\cos^2 x + \\sin^2 x = 1$$\n\nWe substitute this expression directly in place of the term $1$ in the numerator of our LHS expression. This step is key to gaining our first mark **(M1)**:\n\n$$\\text{LHS} = \\frac{(\\cos^2 x + \\sin^2 x) - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":2,"title":"Simplify the Numerator","content":"Now, we simplify the terms in the numerator by grouping the $\\sin^2 x$ terms:\n\n$$\\text{LHS} = \\frac{\\cos^2 x + \\sin^2 x - 2\\sin^2 x}{\\cos x - \\sin x}$$\n\nSubtracting the terms gives:\n\n$$\\text{LHS} = \\frac{\\cos^2 x - \\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":3,"title":"Factor and Cancel","content":"We recognize the numerator as a **difference of two squares** of the form $a^2 - b^2 = (a-b)(a+b)$:\n\n$$\\cos^2 x - \\sin^2 x = (\\cos x - \\sin x)(\\cos x + \\sin x)$$\n\nSubstituting this back into our expression, we have:\n\n$$\\text{LHS} = \\frac{(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\cos x - \\sin x}$$\n\nWe can cancel the common factor $(\\cos x - \\sin x)$ in both the numerator and the denominator, leaving us with the Right-Hand Side **(A1)**:\n\n$$\\text{LHS} = \\cos x + \\sin x$$\n\nSince LHS is shown to equal RHS, the proof is complete."}]},{"id":"double_angle","label":"Double-Angle Formulas","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression:\n\n$$\\text{LHS} = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":1,"title":"Apply double-angle identities","content":"We recognize that the numerator, $1 - 2\\sin^2 x$, is a double-angle formula for $\\cos 2x$:\n\n$$\\cos 2x = 1 - 2\\sin^2 x$$\n\nAnother form of the double-angle formula for $\\cos 2x$ is:\n\n$$\\cos 2x = \\cos^2 x - \\sin^2 x$$\n\nBy equating these, we can rewrite the numerator directly in terms of $\\cos x$ and $\\sin x$:\n\n$$\\text{LHS} = \\frac{\\cos^2 x - \\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":2,"title":"Factorize the numerator","content":"Next, we recognize the numerator $\\cos^2 x - \\sin^2 x$ as a **difference of two squares** of the form $a^2 - b^2 = (a - b)(a + b)$:\n\n$$\\cos^2 x - \\sin^2 x = (\\cos x - \\sin x)(\\cos x + \\sin x)$$\n\nSubstituting this factorization back into our fraction gives:\n\n$$\\text{LHS} = \\frac{(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\cos x - \\sin x}$$"},{"type":"text","order":3,"title":"Cancel and complete","content":"We can now cancel the common factor of $(\\cos x - \\sin x)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\cos x + \\sin x$$\n\nSince the LHS is now equal to the RHS, the proof is complete.\n\n### Markscheme Breakdown\n- **M1**: Awarded for recognizing $1 - 2\\sin^2 x = \\cos 2x$ and converting it to $\\cos^2 x - \\sin^2 x$.\n- **A1**: Awarded for factorizing the numerator, canceling the denominator, and correctly reaching the RHS."}]},{"id":"reverse_multiplication","label":"Working from Right-to-Left","steps":[{"type":"text","order":0,"title":"Understand the approach","content":"To prove the identity using the **Right-to-Left** approach, we want to establish a link between the right-hand side (RHS) and the left-hand side (LHS).\n\nLet's start by multiplying the RHS expression by the denominator of the LHS to see if it simplifies to the numerator of the LHS:\n\n$$\\text{RHS} \\times \\text{Denominator of LHS} = (\\cos x + \\sin x)(\\cos x - \\sin x)$$"},{"type":"text","order":1,"title":"Apply difference of squares","content":"We expand this product using the difference of squares identity, $(a+b)(a-b) = a^2 - b^2$:\n\n$$(\\cos x + \\sin x)(\\cos x - \\sin x) = \\cos^2 x - \\sin^2 x$$"},{"type":"text","order":2,"title":"Use Pythagorean identity","content":"Recall the fundamental Pythagorean identity:\n\n$$\\cos^2 x + \\sin^2 x = 1 \\implies \\cos^2 x = 1 - \\sin^2 x$$\n\nSubstituting this for $\\cos^2 x$ in our expression gives:\n\n$$\\cos^2 x - \\sin^2 x = (1 - \\sin^2 x) - \\sin^2 x$$\n\nSimplifying this expression yields:\n\n$$\\cos^2 x - \\sin^2 x = 1 - 2\\sin^2 x$$\n\nNotice that this is exactly the numerator of our LHS!"},{"type":"text","order":3,"title":"Complete the proof","content":"We have shown that:\n\n$$(\\cos x + \\sin x)(\\cos x - \\sin x) = 1 - 2\\sin^2 x$$\n\nNow, dividing both sides by $(\\cos x - \\sin x)$ isolates the RHS expression:\n\n$$\\cos x + \\sin x = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$\n\nThis is exactly LHS = RHS, which completes our proof.\n\n### Markscheme Breakdown\n- **M1**: Awarded for substituting $\\cos^2 x = 1 - \\sin^2 x$ (or equivalent identity work).\n- **A1**: Awarded for successfully completing the simplification to show both sides are equal."}]}],"generatedAt":"2026-06-09T04:15:43.542Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2b72a823-4676-4a0e-933d-bdced993c272:pythagorean_substitution","label":"Pythagorean Identity Substitution","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression:\n\n$$\\text{LHS} = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":1,"title":"Substitute the Pythagorean Identity","content":"Recall the fundamental Pythagorean identity:\n\n$$\\cos^2 x + \\sin^2 x = 1$$\n\nWe substitute this expression directly in place of the term $1$ in the numerator of our LHS expression. This step is key to gaining our first mark **(M1)**:\n\n$$\\text{LHS} = \\frac{(\\cos^2 x + \\sin^2 x) - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":2,"title":"Simplify the Numerator","content":"Now, we simplify the terms in the numerator by grouping the $\\sin^2 x$ terms:\n\n$$\\text{LHS} = \\frac{\\cos^2 x + \\sin^2 x - 2\\sin^2 x}{\\cos x - \\sin x}$$\n\nSubtracting the terms gives:\n\n$$\\text{LHS} = \\frac{\\cos^2 x - \\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":3,"title":"Factor and Cancel","content":"We recognize the numerator as a **difference of two squares** of the form $a^2 - b^2 = (a-b)(a+b)$:\n\n$$\\cos^2 x - \\sin^2 x = (\\cos x - \\sin x)(\\cos x + \\sin x)$$\n\nSubstituting this back into our expression, we have:\n\n$$\\text{LHS} = \\frac{(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\cos x - \\sin x}$$\n\nWe can cancel the common factor $(\\cos x - \\sin x)$ in both the numerator and the denominator, leaving us with the Right-Hand Side **(A1)**:\n\n$$\\text{LHS} = \\cos x + \\sin x$$\n\nSince LHS is shown to equal RHS, the proof is complete."}]},{"id":"2b72a823-4676-4a0e-933d-bdced993c272:double_angle","label":"Double-Angle Formulas","steps":[{"type":"text","order":0,"title":"Start with the LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression:\n\n$$\\text{LHS} = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":1,"title":"Apply double-angle identities","content":"We recognize that the numerator, $1 - 2\\sin^2 x$, is a double-angle formula for $\\cos 2x$:\n\n$$\\cos 2x = 1 - 2\\sin^2 x$$\n\nAnother form of the double-angle formula for $\\cos 2x$ is:\n\n$$\\cos 2x = \\cos^2 x - \\sin^2 x$$\n\nBy equating these, we can rewrite the numerator directly in terms of $\\cos x$ and $\\sin x$:\n\n$$\\text{LHS} = \\frac{\\cos^2 x - \\sin^2 x}{\\cos x - \\sin x}$$"},{"type":"text","order":2,"title":"Factorize the numerator","content":"Next, we recognize the numerator $\\cos^2 x - \\sin^2 x$ as a **difference of two squares** of the form $a^2 - b^2 = (a - b)(a + b)$:\n\n$$\\cos^2 x - \\sin^2 x = (\\cos x - \\sin x)(\\cos x + \\sin x)$$\n\nSubstituting this factorization back into our fraction gives:\n\n$$\\text{LHS} = \\frac{(\\cos x - \\sin x)(\\cos x + \\sin x)}{\\cos x - \\sin x}$$"},{"type":"text","order":3,"title":"Cancel and complete","content":"We can now cancel the common factor of $(\\cos x - \\sin x)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\cos x + \\sin x$$\n\nSince the LHS is now equal to the RHS, the proof is complete.\n\n### Markscheme Breakdown\n- **M1**: Awarded for recognizing $1 - 2\\sin^2 x = \\cos 2x$ and converting it to $\\cos^2 x - \\sin^2 x$.\n- **A1**: Awarded for factorizing the numerator, canceling the denominator, and correctly reaching the RHS."}]},{"id":"2b72a823-4676-4a0e-933d-bdced993c272:reverse_multiplication","label":"Working from Right-to-Left","steps":[{"type":"text","order":0,"title":"Understand the approach","content":"To prove the identity using the **Right-to-Left** approach, we want to establish a link between the right-hand side (RHS) and the left-hand side (LHS).\n\nLet's start by multiplying the RHS expression by the denominator of the LHS to see if it simplifies to the numerator of the LHS:\n\n$$\\text{RHS} \\times \\text{Denominator of LHS} = (\\cos x + \\sin x)(\\cos x - \\sin x)$$"},{"type":"text","order":1,"title":"Apply difference of squares","content":"We expand this product using the difference of squares identity, $(a+b)(a-b) = a^2 - b^2$:\n\n$$(\\cos x + \\sin x)(\\cos x - \\sin x) = \\cos^2 x - \\sin^2 x$$"},{"type":"text","order":2,"title":"Use Pythagorean identity","content":"Recall the fundamental Pythagorean identity:\n\n$$\\cos^2 x + \\sin^2 x = 1 \\implies \\cos^2 x = 1 - \\sin^2 x$$\n\nSubstituting this for $\\cos^2 x$ in our expression gives:\n\n$$\\cos^2 x - \\sin^2 x = (1 - \\sin^2 x) - \\sin^2 x$$\n\nSimplifying this expression yields:\n\n$$\\cos^2 x - \\sin^2 x = 1 - 2\\sin^2 x$$\n\nNotice that this is exactly the numerator of our LHS!"},{"type":"text","order":3,"title":"Complete the proof","content":"We have shown that:\n\n$$(\\cos x + \\sin x)(\\cos x - \\sin x) = 1 - 2\\sin^2 x$$\n\nNow, dividing both sides by $(\\cos x - \\sin x)$ isolates the RHS expression:\n\n$$\\cos x + \\sin x = \\frac{1 - 2\\sin^2 x}{\\cos x - \\sin x}$$\n\nThis is exactly LHS = RHS, which completes our proof.\n\n### Markscheme Breakdown\n- **M1**: Awarded for substituting $\\cos^2 x = 1 - \\sin^2 x$ (or equivalent identity work).\n- **A1**: Awarded for successfully completing the simplification to show both sides are equal."}]}],"generatedAt":"2026-06-09T04:15:56.845Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"934c88ce-aae4-46e7-8f4e-4af676f41791","revisionId":"a55d597d-e871-4976-b388-129be38b081a","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"4f2efc02-2b6f-4298-af8b-24851da426b5","label":"a","content":"Prove the identity:\n\n$$\\frac{\\cos 4A}{\\cos 2A + \\sin 2A} \\equiv \\cos 2A - \\sin 2A$$","markscheme":"- Substitutes $\\cos 4A = \\cos^2 2A - \\sin^2 2A$ M1\n- Factorises numerator as $(\\cos 2A - \\sin 2A)(\\cos 2A + \\sin 2A)$ and cancels $(\\cos 2A + \\sin 2A)$ to reach RHS A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Apply the double angle identity to cos(4A) to get $\\cos^2 2A - \\sin^2 2A$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly factorise and cancel the denominator terms to obtain the RHS.","code":"A","acceptable":["LHS = (cos 2A - sin 2A)(cos 2A + sin 2A) / (cos 2A + sin 2A) = cos 2A - sin 2A"]}],"finalAnswers":["\\cos 2A - \\sin 2A"],"generatedAt":"2026-05-06T14:03:54.761Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"lhs-to-rhs","label":"LHS to RHS Expansion","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression and simplify it step-by-step to reach the Right-Hand Side (RHS):\n\n$$\\text{LHS} = \\frac{\\cos 4A}{\\cos 2A + \\sin 2A}$$,"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\cos(2\\theta) = \\cos^2\\theta - \\sin^2\\theta$"},{"latex":"$$\\cos(\\htmlData{anchor=angle-double}{4A}) = \\cos^2(\\htmlData{anchor=angle-half-1}{2A}) - \\sin^2(\\htmlData{anchor=angle-half-2}{2A})$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=num-identity}{\\cos^2 2A - \\sin^2 2A}}{\\cos 2A + \\sin 2A}$"}],"highlights":[{"color":"sky","style":"box","anchor":"num-identity"}],"connections":[]},"order":1,"title":"Apply double-angle identity","description":"Recall the double-angle identity for cosine: $$\\cos(2\\theta) = \\cos^2\\theta - \\sin^2\\theta$$ Let's apply this identity by substituting $\\theta = 2A$ to rewrite the numerator $\\cos 4A$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 - y^2 = (x - y)(x + y)$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=factored-num}{\\color{@purple}{(\\cos 2A - \\sin 2A)(\\cos 2A + \\sin 2A)}}}{\\htmlData{anchor=denom}{\\cos 2A + \\sin 2A}}$"}],"highlights":[{"color":"purple","style":"underline","anchor":"factored-num"}],"connections":[]},"order":2,"title":"Factorise the numerator","description":"The numerator $\\cos^2 2A - \\sin^2 2A$ is a difference of two squares. We can factorise it using the algebraic pattern $x^2 - y^2 = (x - y)(x + y)$."},{"type":"text","order":3,"title":"Cancel common factors","content":"Now, we can cancel the common factor of $(\\cos 2A + \\sin 2A)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\frac{(\\cos 2A - \\sin 2A)\\cancel{(\\cos 2A + \\sin 2A)}}{\\cancel{\\cos 2A + \\sin 2A}}$$\n\nThis leaves us with:\n\n$$\\text{LHS} = \\cos 2A - \\sin 2A = \\text{RHS}$$\n\nThis completes the proof. (This satisfies the **M1** mark for substituting the double-angle identity and the **A1** mark for factorising and cancelling to reach the final simplified expression)."}]},{"id":"rhs-to-lhs","label":"RHS to LHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using RHS to LHS manipulation, we start with the Right-Hand Side (RHS):\n\n$$\\text{RHS} = \\cos 2A - \\sin 2A$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\htmlData{anchor=rhs-expr}{\\color{@sky}{\\cos 2A - \\sin 2A}}$"},{"latex":"$$\\text{RHS} = \\frac{(\\htmlData{anchor=rhs-sub}{\\color{@sky}{\\cos 2A - \\sin 2A}})(\\htmlData{anchor=conj-term}{\\color{@amber}{\\cos 2A + \\sin 2A}})}{\\htmlData{anchor=denom-term}{\\color{@amber}{\\cos 2A + \\sin 2A}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rhs-expr"},{"color":"sky","style":"text-color","anchor":"rhs-sub"},{"color":"amber","style":"text-color","anchor":"conj-term"},{"color":"amber","style":"text-color","anchor":"denom-term"}],"connections":[{"to":"rhs-sub","from":"rhs-expr","color":"sky","label":"RHS","style":"solid"}]},"order":1,"title":"Multiply and divide","description":"To introduce the fractional structure of the LHS, we multiply and divide the RHS expression by $(\\cos 2A + \\sin 2A)$."},{"type":"text","order":2,"title":"Expand the numerator","content":"The numerator is now in the form of a difference of two squares:\n\n$$(x - y)(x + y) = x^2 - y^2$$\n\nSubstituting $x = \\cos 2A$ and $y = \\sin 2A$, we expand the numerator:\n\n$$\\text{RHS} = \\frac{\\cos^2 2A - \\sin^2 2A}{\\cos 2A + \\sin 2A}$$"},{"type":"text","order":3,"title":"Apply double-angle identity","content":"Recall the double-angle identity for cosine:\n\n$$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$\n\nLetting $\\theta = 2A$, we get:\n\n$$\\cos 4A = \\cos^2 2A - \\sin^2 2A$$\n\nSubstituting this back into the numerator of our expression yields:\n\n$$\\text{RHS} = \\frac{\\cos 4A}{\\cos 2A + \\sin 2A}$$\n\nThis is identical to the Left-Hand Side (LHS), completing our proof:\n\n$$\\text{RHS} \\equiv \\text{LHS}$$"}]}],"generatedAt":"2026-06-09T07:41:10.830Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4f2efc02-2b6f-4298-af8b-24851da426b5:lhs-to-rhs","label":"LHS to RHS Expansion","steps":[{"type":"text","order":0,"title":"Start with LHS","content":"To prove the identity, we will start with the Left-Hand Side (LHS) of the expression and simplify it step-by-step to reach the Right-Hand Side (RHS):\n\n$$\\text{LHS} = \\frac{\\cos 4A}{\\cos 2A + \\sin 2A}$$,"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\cos(2\\theta) = \\cos^2\\theta - \\sin^2\\theta$"},{"latex":"$$\\cos(\\htmlData{anchor=angle-double}{4A}) = \\cos^2(\\htmlData{anchor=angle-half-1}{2A}) - \\sin^2(\\htmlData{anchor=angle-half-2}{2A})$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=num-identity}{\\cos^2 2A - \\sin^2 2A}}{\\cos 2A + \\sin 2A}$"}],"highlights":[{"color":"sky","style":"box","anchor":"num-identity"}],"connections":[]},"order":1,"title":"Apply double-angle identity","description":"Recall the double-angle identity for cosine: $$\\cos(2\\theta) = \\cos^2\\theta - \\sin^2\\theta$$ Let's apply this identity by substituting $\\theta = 2A$ to rewrite the numerator $\\cos 4A$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2 - y^2 = (x - y)(x + y)$"},{"latex":"$$\\text{LHS} = \\frac{\\htmlData{anchor=factored-num}{\\color{@purple}{(\\cos 2A - \\sin 2A)(\\cos 2A + \\sin 2A)}}}{\\htmlData{anchor=denom}{\\cos 2A + \\sin 2A}}$"}],"highlights":[{"color":"purple","style":"underline","anchor":"factored-num"}],"connections":[]},"order":2,"title":"Factorise the numerator","description":"The numerator $\\cos^2 2A - \\sin^2 2A$ is a difference of two squares. We can factorise it using the algebraic pattern $x^2 - y^2 = (x - y)(x + y)$."},{"type":"text","order":3,"title":"Cancel common factors","content":"Now, we can cancel the common factor of $(\\cos 2A + \\sin 2A)$ from both the numerator and the denominator:\n\n$$\\text{LHS} = \\frac{(\\cos 2A - \\sin 2A)\\cancel{(\\cos 2A + \\sin 2A)}}{\\cancel{\\cos 2A + \\sin 2A}}$$\n\nThis leaves us with:\n\n$$\\text{LHS} = \\cos 2A - \\sin 2A = \\text{RHS}$$\n\nThis completes the proof. (This satisfies the **M1** mark for substituting the double-angle identity and the **A1** mark for factorising and cancelling to reach the final simplified expression)."}]},{"id":"4f2efc02-2b6f-4298-af8b-24851da426b5:rhs-to-lhs","label":"RHS to LHS Manipulation","steps":[{"type":"text","order":0,"title":"Start with the RHS","content":"To prove the identity using RHS to LHS manipulation, we start with the Right-Hand Side (RHS):\n\n$$\\text{RHS} = \\cos 2A - \\sin 2A$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{RHS} = \\htmlData{anchor=rhs-expr}{\\color{@sky}{\\cos 2A - \\sin 2A}}$"},{"latex":"$$\\text{RHS} = \\frac{(\\htmlData{anchor=rhs-sub}{\\color{@sky}{\\cos 2A - \\sin 2A}})(\\htmlData{anchor=conj-term}{\\color{@amber}{\\cos 2A + \\sin 2A}})}{\\htmlData{anchor=denom-term}{\\color{@amber}{\\cos 2A + \\sin 2A}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rhs-expr"},{"color":"sky","style":"text-color","anchor":"rhs-sub"},{"color":"amber","style":"text-color","anchor":"conj-term"},{"color":"amber","style":"text-color","anchor":"denom-term"}],"connections":[{"to":"rhs-sub","from":"rhs-expr","color":"sky","label":"RHS","style":"solid"}]},"order":1,"title":"Multiply and divide","description":"To introduce the fractional structure of the LHS, we multiply and divide the RHS expression by $(\\cos 2A + \\sin 2A)$."},{"type":"text","order":2,"title":"Expand the numerator","content":"The numerator is now in the form of a difference of two squares:\n\n$$(x - y)(x + y) = x^2 - y^2$$\n\nSubstituting $x = \\cos 2A$ and $y = \\sin 2A$, we expand the numerator:\n\n$$\\text{RHS} = \\frac{\\cos^2 2A - \\sin^2 2A}{\\cos 2A + \\sin 2A}$$"},{"type":"text","order":3,"title":"Apply double-angle identity","content":"Recall the double-angle identity for cosine:\n\n$$\\cos 2\\theta = \\cos^2 \\theta - \\sin^2 \\theta$$\n\nLetting $\\theta = 2A$, we get:\n\n$$\\cos 4A = \\cos^2 2A - \\sin^2 2A$$\n\nSubstituting this back into the numerator of our expression yields:\n\n$$\\text{RHS} = \\frac{\\cos 4A}{\\cos 2A + \\sin 2A}$$\n\nThis is identical to the Left-Hand Side (LHS), completing our proof:\n\n$$\\text{RHS} \\equiv \\text{LHS}$$"}]}],"generatedAt":"2026-06-09T07:41:22.307Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__aqa","summaries":[{"id":"e0cd8580-436f-4f08-8dee-284b953c607a","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"f4923309-8d0b-4a75-bdf8-bf6885937510","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"f72cfefd-990a-475c-8fde-0ec350de12d9","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"327076de-b61b-400a-bde2-937a30472d46","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"78a118f7-2344-4c79-b026-e5f3ada312c2","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"c982acf5-de7a-460e-bb95-d152177fb4e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"536015ff-d008-4174-9c6c-839c46217f8e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"775582c4-55c4-4dc3-8c5c-c0a63c63dcaa","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"934c88ce-aae4-46e7-8f4e-4af676f41791","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"536ab148-9e3e-429a-8f23-80ed01ef8557","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"a2eec4d2-89c8-415c-8c8b-6b1e523138f7","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"cd1dffd5-3636-4a2e-a756-86ef66e4090d","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"006483fe-fe7e-4cc4-864d-fab5f152bf33","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0523befd-5c19-49fa-8e31-d8a34244e35e","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"11e7f755-bfdb-461a-bc4f-9c1680670249","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"1fde01cc-33ca-4271-a99a-2994b71cce0f","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"21390a73-0ded-4be9-9fc7-4562b609cfb0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"21c02aed-c565-49a6-88c1-520b965b5fcd","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"2423687b-ba39-452d-8863-34deb93dac6c","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"337be6c9-7358-491a-a890-000198ede2d4","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"3db9f5cd-0c63-447a-a354-0c61624681b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"4915fd73-abcd-48ff-973d-c8f1edf5fdd7","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"505bffff-22b1-4097-9302-fa5a280cb6e2","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"50e9b35b-5950-4c9f-9fce-dfdb9de941a3","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"6353b0eb-3a53-4d5d-b1e7-14fe34fe647c","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"68a4587e-20bd-414a-9532-c9dd05cb5862","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"728d4ae9-23d0-424a-8951-12bc6cae52a5","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"7a783d21-a95d-4b5d-9a49-84ac3e4e9b0b","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"7fd5d09a-84b7-4cef-9514-e65fe32d4c3a","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"8e1db89f-bb68-4109-9390-5447ba387392","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"96a08353-fb4d-46d3-8988-dfa69d9bf244","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"ab79c082-1eb6-4721-8714-72817b51f94c","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"af834080-99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Proofs with trigonometric identities","topicId":"de74cbad-b351-4549-a303-5b583fd28577","topicSidebar":"$L3d","topicSlug":"proofs-with-trigonometric-identities","topicTitle":"1.8.8 Proofs with trigonometric identities"}]}],"$L3e","$L3f"]}]

1.8.8 Proofs with trigonometric identities

Question 2

1.8.8 Proofs with trigonometric identities Questions

1.8.8 Proofs with trigonometric identities

Question 2

1.8.8 Proofs with trigonometric identities Questions