Q.Integrate the following function: $\sin 4x \sin 8x$

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\left(1 + \cos\dfrac{\pi}{8}\right)\left(1 + \cos\dfrac{3\pi}{8}\right)\left(1 + \cos\dfrac{5\pi}{8}\right)\left(1 + \cos\dfrac{7\pi}{8}\right) =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{1+\sqrt{2}}{\sqrt{2}}$

   ›Reveal solutionSolution

   Tests the trick of pairing supplementary-angle cosine terms to turn $(1+\cos\theta)(1-\cos\theta)$ into $\sin^2\theta$. Answer: $\frac18$.

   Concept and Intuition

   The four angles $\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}$ are symmetric about $\frac{\pi}{2}$: the last two are supplements of the first two ($\frac{5\pi}{8}=\pi-\frac{3\pi}{8}$, $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$). Since $\cos(\pi-\theta)=-\cos\theta$, the four factors regroup into two conjugate pairs $(1+\cos\theta)(1-\cos\theta)$, which collapse via the difference-of-squares identity to $\sin^2\theta$.

   Step-by-Step Solution

   1. $\cos\frac{5\pi}{8}=\cos\left(\pi-\frac{3\pi}{8}\right)=-\cos\frac{3\pi}{8}$, and $\cos\frac{7\pi}{8}=\cos\left(\pi-\frac{\pi}{8}\right)=-\cos\frac{\pi}{8}$.
   2. The product becomes $\left(1+\cos\tfrac{\pi}{8}\right)\left(1-\cos\tfrac{\pi}{8}\right)\left(1+\cos\tfrac{3\pi}{8}\right)\left(1-\cos\tfrac{3\pi}{8}\right)=\left(1-\cos^2\tfrac{\pi}{8}\right)\left(1-\cos^2\tfrac{3\pi}{8}\right)=\sin^2\tfrac{\pi}{8}\sin^2\tfrac{3\pi}{8}.$ …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The sum of all the values of $\theta \in (0, 2\pi)$ satisfying the equation $\sin\theta + 3\cos 2\theta + \sin 3\theta = \cos\theta + 3\sin 2\theta + \cos 3\theta$ is (A) $\dfrac{5\pi}{8}$ (B) $\dfrac{13\pi}{8}$ (C) $\dfrac{32\pi}{8}$ (D) $\dfrac{28\pi}{8}$

   ›Reveal solutionSolution

   Regroup the six-term trig equation into a product of two factors; one factor never vanishes, leaving a simple $\tan2\theta=1$ condition.

   Concept and Intuition

   Six scattered trig terms are a strong hint to pair them so sum-to-product identities apply. Pairing $\sin\theta$ with $\sin3\theta$ (average angle $2\theta$) and $\cos\theta$ with $\cos3\theta$ (also average $2\theta$) turns four terms into two terms sharing the common factor $\cos\theta$, which then factors neatly against the remaining $\cos2\theta,\sin2\theta$ pair.

   Step-by-Step Solution

   1. Rearrange the equation: $(\sin\theta+\sin3\theta) - (\cos\theta+\cos3\theta) + 3(\cos2\theta - \sin2\theta) = 0$.
   2. Sum-to-product: $\sin\theta+\sin3\theta = 2\sin2\theta\cos\theta$, and $\cos\theta+\cos3\theta = 2\cos2\theta\cos\theta$.
   3. Substitute: $2\sin2\theta\cos\theta - 2\cos2\theta\cos\theta + 3\cos2\theta - 3\sin2\theta = 0$.
   4. Group: $2\cos\theta(\sin2\theta-\cos2\theta) - 3(\sin2\theta-\cos2\theta) = 0$.
   5. Factor: $(\sin2\theta-\cos2\theta)(2\cos\theta-3)=0$.
   6. $2\cos\theta-3=0 \Rightarrow \cos\theta=3/2$ — impossible, discard. …

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $x \neq (2n+1)\dfrac{\pi}{4}$, then the general solution of $\cos x + \cos 3x = \sin x + \sin 3x$ is (A) $n\pi + \dfrac{\pi}{8}$ (B) $n\pi \pm \dfrac{\pi}{8}$ (C) $\dfrac{n\pi}{2} \pm \dfrac{\pi}{8}$ (D) $\dfrac{n\pi}{2} + \dfrac{\pi}{8}$

   ›Reveal solutionSolution

   Convert both sides to a product form using sum-to-product identities, factor out $\cos x$, and solve $\tan2x=1$ for the general solution — the given domain restriction is exactly what makes this branch well-defined. Answer: $x=\dfrac{n\pi}{2}+\dfrac\pi8$.

   Concept and Intuition

   Sum-to-product identities turn a sum of cosines/sines into a product, which is powerful for solving trig equations because a product equal to zero splits into simpler independent equations. Here both sums share the common factor $\cos x$, and the interesting equation reduces to a basic $\tan2x=1$.

   Step-by-Step Solution

   1. $\cos x+\cos3x=2\cos\!\left(\dfrac{x+3x}{2}\right)\cos\!\left(\dfrac{3x-x}{2}\right)=2\cos2x\cos x$.
   2. $\sin x+\sin3x=2\sin\!\left(\dfrac{x+3x}{2}\right)\cos\!\left(\dfrac{3x-x}{2}\right)=2\sin2x\cos x$.
   3. The equation $\cos x+\cos3x=\sin x+\sin3x$ becomes $2\cos2x\cos x=2\sin2x\cos x$, i.e. $2\cos x(\cos2x-\sin2x)=0$.
   4. The condition $x\ne(2n+1)\pi/4$ is precisely the condition $\cos2x\ne0$, which is exactly what's needed to safely write $\tan2x=\dfrac{\sin2x}{\cos2x}$ from $\cos2x=\sin2x$ — signalling that the intended branch to solve is $\cos2x-\sin2x=0$. …

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4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$\sin\frac{\pi}{16} \sin\frac{3\pi}{16} \sin\frac{5\pi}{16} \sin\frac{7\pi}{16} =$ (A) $\frac{\sqrt{2}}{16}$ (B) $\frac{1}{8}$ (C) $\frac{1}{16}$ (D) $\frac{\sqrt{2}}{32}$

   ›Reveal solutionSolution

   Using complementary-angle pairing and the double-angle identity, the product collapses to $\frac{\sqrt2}{16}$.

   Concept and Intuition

   When angles inside a product of sines add up to $\pi/2$, pairing them as sine and cosine of the same angle lets us use $\sin\theta\cos\theta=\frac12\sin2\theta$ repeatedly to collapse the whole product.

   Step-by-Step Solution

   1. Note $\frac{\pi}{16}+\frac{7\pi}{16}=\frac{\pi}{2}$ so $\sin\frac{7\pi}{16}=\cos\frac{\pi}{16}$.
   2. Note $\frac{3\pi}{16}+\frac{5\pi}{16}=\frac{\pi}{2}$ so $\sin\frac{5\pi}{16}=\cos\frac{3\pi}{16}$.
   3. The product becomes $\left(\sin\frac{\pi}{16}\cos\frac{\pi}{16}\right)\left(\sin\frac{3\pi}{16}\cos\frac{3\pi}{16}\right)$.
   4. $=\frac12\sin\frac{\pi}{8}\cdot\frac12\sin\frac{3\pi}{8} = \frac14\sin\frac{\pi}{8}\sin\frac{3\pi}{8}$. …

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5. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $P+Q+R=\dfrac{\pi}{4}$, then $\cos\left(\dfrac{\pi}{8} - P\right) + \cos\left(\dfrac{\pi}{8} - Q\right) + \cos\left(\dfrac{\pi}{8} - R\right) =$ (A) $4\cos\dfrac{P}{2}\cos\dfrac{Q}{2}\cos\dfrac{R}{2} - \cos\dfrac{\pi}{8}$ (B) $4\cos\dfrac{P}{2}\cos\dfrac{Q}{2}\sin\dfrac{R}{2} + \cos\dfrac{\pi}{8}$ (C) $4\sin\dfrac{P}{2}\sin\dfrac{Q}{2}\sin\dfrac{R}{2} - \cos\dfrac{\pi}{8}$ (D) $4\sin\dfrac{P}{2}\cos\dfrac{Q}{2}\sin\dfrac{R}{2} + \cos\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   This tests the sum-to-product technique for three angles constrained by $P+Q+R=\pi/4$; the answer is $4\cos\frac{P}{2}\cos\frac{Q}{2}\cos\frac{R}{2}-\cos\frac{\pi}{8}$.

   Concept and Intuition

   Whenever three angles are linked by a linear constraint like $P+Q+R=$ constant, the standard move is to combine two of the cosine terms via sum-to-product so that the constraint eliminates one variable, then use product-to-sum again to fold the remaining pieces into a symmetric product. This mirrors the classical identity for a triangle ($A+B+C=\pi \Rightarrow \cos A+\cos B+\cos C = 1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$), just with a different target constant ($\pi/4$ instead of $\pi$).

   Step-by-Step Solution

   1. Combine the first two terms using $\cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ with $C=\frac{\pi}{8}-P,\ D=\frac{\pi}{8}-Q$:

   $\cos\left(\tfrac{\pi}{8}-P\right)+\cos\left(\tfrac{\pi}{8}-Q\right)=2\cos\left(\tfrac{\pi}{8}-\tfrac{P+Q}{2}\right)\cos\left(\tfrac{Q-P}{2}\right).$

   2. Since $P+Q=\frac{\pi}{4}-R$, we get $\frac{\pi}{8}-\frac{P+Q}{2}=\frac{\pi}{8}-\frac{\pi}{8}+\frac{R}{2}=\frac{R}{2}$. So the pair sums to $2\cos\frac{R}{2}\cos\left(\frac{Q-P}{2}\right)$.
   3. Use product-to-sum on $\cos\frac{P}{2}\cos\frac{Q}{2}$: $2\cos\frac{P}{2}\cos\frac{Q}{2}=\cos\left(\frac{Q-P}{2}\right)+\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$ (using $\frac{P+Q}{2}=\frac{\pi}{8}-\frac{R}{2}$ again). So $\cos\left(\frac{Q-P}{2}\right)=2\cos\frac{P}{2}\cos\frac{Q}{2}-\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$.
   4. Substitute: $2\cos\frac{R}{2}\cos\left(\frac{Q-P}{2}\right)=4\cos\frac{P}{2}\cos\frac{Q}{2}\cos\frac{R}{2}-2\cos\frac{R}{2}\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$.
   5. Expand the last product using $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ with $A=\frac{R}{2},B=\frac{\pi}{8}-\frac{R}{2}$: it equals $\cos\left(\frac{\pi}{8}-R\right)+\cos\frac{\pi}{8}$. …

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6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The general solution of the equation $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$ is (A) $n\pi + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (B) $\dfrac{n\pi}{2} + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (C) $(-1)^n\dfrac{n\pi}{2} + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (D) $2n\pi + \cos^{-1}\dfrac{3}{2}, n \in \mathbb{Z}$

   ›Reveal solutionSolution

   A sum-to-product regrouping turns the equation into $(2\cos x-3)(\sin2x-\cos2x)=0$; the first factor never vanishes, so the general solution is $x=\frac{n\pi}{2}+\frac{\pi}{8}$.

   Concept and Intuition

   When an equation mixes $x$, $2x$, $3x$ terms, grouping the outer terms ($x$ and $3x$) via sum-to-product often exposes a common factor with the middle term — turning a transcendental-looking equation into a simple factorable one.

   Step-by-Step Solution

   1. $\sin x+\sin 3x = 2\sin 2x\cos x$ and $\cos x + \cos 3x = 2\cos 2x\cos x$ (sum-to-product, since the mean angle is $2x$ and half-difference is $x$).
   2. LHS of the given equation $=2\sin2x\cos x - 3\sin2x = \sin2x(2\cos x - 3)$.
   3. RHS $=2\cos2x\cos x - 3\cos2x = \cos2x(2\cos x-3)$.
   4. Equation: $\sin2x(2\cos x-3)=\cos2x(2\cos x-3)$, i.e. $(2\cos x-3)(\sin2x-\cos2x)=0$.
   5. $2\cos x = 3$ has no real solution (cosine is bounded by 1), so we must have $\sin2x=\cos2x$, i.e. $\tan2x=1$. …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$4\cos\dfrac{\pi}{7}\cos\dfrac{\pi}{5}\cos\dfrac{2\pi}{7}\cos\dfrac{2\pi}{5}\cos\dfrac{4\pi}{7} =$ (A) $-\dfrac{1}{8}$ (B) $\dfrac{1}{32}$ (C) $-\dfrac{1}{32}$ (D) $\dfrac{1}{8}$

   ›Reveal solutionSolution

   Splitting the five-factor product into two known standard identities — one for sevenths, one for fifths of $\pi$ — gives $-\dfrac18$ directly.

   Concept and Intuition

   Two classical product identities make this tractable: $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}=\frac18$ (and since $\cos\frac{4\pi}{7}=-\cos\frac{3\pi}{7}$, a variant with $\frac{4\pi}{7}$ instead of $\frac{3\pi}{7}$ flips sign), and $\cos36^\circ\cos72^\circ=\frac14$ (using the exact golden-ratio values of $\cos36^\circ,\cos72^\circ$).

   Step-by-Step Solution

   1. Known identity: $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac18$.
   2. Since $\dfrac{4\pi}{7}=\pi-\dfrac{3\pi}{7}$, we have $\cos\dfrac{4\pi}{7}=-\cos\dfrac{3\pi}{7}$.
   3. So $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{4\pi}{7}=-\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=-\dfrac18$. …

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8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.$\left(1+\cos\dfrac{\pi}{8}\right)\left(1+\cos\dfrac{2\pi}{8}\right)\left(1+\cos\dfrac{3\pi}{8}\right)\left(1+\cos\dfrac{4\pi}{8}\right)\ldots\left(1+\cos\dfrac{7\pi}{8}\right)=$ (A) $\dfrac{1}{16}$ (B) $\dfrac{1}{64}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{3}{64}$

   ›Reveal solutionSolution

   Convert each factor $1+\cos\theta$ into $2\cos^2(\theta/2)$, then apply the classical cosine-product identity $\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt n}{2^{n-1}}$.

   Concept and Intuition

   The half-angle identity $1+\cos\theta=2\cos^2(\theta/2)$ converts the whole product of seven $(1+\cos)$ factors into a product of squared cosines with angles forming an arithmetic-like sequence $\frac{\pi}{16},\frac{2\pi}{16},\dots,\frac{7\pi}{16}$. This matches the well-known identity $\prod_{k=1}^{n-1}\cos\left(\dfrac{k\pi}{2n}\right)=\dfrac{\sqrt n}{2^{n-1}}$ with $2n=16$, i.e. $n=8$.

   Step-by-Step Solution

   1. Each factor: $1+\cos\dfrac{k\pi}{8} = 2\cos^2\left(\dfrac{k\pi}{16}\right)$ for $k=1,\dots,7$.
   2. Product $= 2^7\displaystyle\prod_{k=1}^{7}\cos^2\left(\dfrac{k\pi}{16}\right)$.
   3. Apply the identity $\displaystyle\prod_{k=1}^{n-1}\cos\left(\dfrac{k\pi}{2n}\right)=\dfrac{\sqrt n}{2^{n-1}}$ with $n=8$: $\displaystyle\prod_{k=1}^{7}\cos\left(\dfrac{k\pi}{16}\right) = \dfrac{\sqrt8}{2^7} = \dfrac{2\sqrt2}{128}=\dfrac{\sqrt2}{64}$. …

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9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $f(x) = \sin x \cdot \sin 2x \cdot \sin 3x$ and $f''(x) = a(\sin bx) + c(\sin dx) + e(\sin kx)$, then the value of $(a + c + e) - (b + d + k)$ equals ______ (A) $8$ (B) $-8$ (C) $16$ (D) $12$

   ›Reveal solutionSolution

   Tests reducing a triple-sine product to a sum of single sines via product-to-sum identities, then differentiating twice and reading off coefficients.

   Concept and Intuition

   Products of sines are hard to differentiate repeatedly in their original form, but converting them into a sum of sines (using $\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]$ repeatedly) turns the problem into differentiating simple sine terms, each of which just picks up a power of its own frequency (with alternating sign) under repeated differentiation.

   Step-by-Step Solution

   1. First combine $\sin2x\sin3x=\dfrac12[\cos(3x-2x)-\cos(3x+2x)]=\dfrac12[\cos x-\cos5x]$.
   2. So $f(x)=\sin x\cdot\dfrac12[\cos x-\cos5x]=\dfrac12[\sin x\cos x-\sin x\cos5x]$.
   3. Use $\sin x\cos x=\dfrac12\sin2x$ and $\sin x\cos5x=\dfrac12[\sin(x+5x)+\sin(x-5x)]=\dfrac12[\sin6x-\sin4x]$.
   4. So $f(x)=\dfrac12\left[\dfrac12\sin2x-\dfrac12(\sin6x-\sin4x)\right]=\dfrac14[\sin2x+\sin4x-\sin6x]$.
   5. Differentiate once: $f'(x)=\dfrac14[2\cos2x+4\cos4x-6\cos6x]=\dfrac12\cos2x+\cos4x-\dfrac32\cos6x$.
   6. Differentiate again: $f''(x)=\dfrac12(-2\sin2x)+(-4\sin4x)-\dfrac32(-6\sin6x)=-\sin2x-4\sin4x+9\sin6x$. …

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10. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.$\sin\dfrac{8\pi}{9}\, \sin\dfrac{7\pi}{9}\, \sin\dfrac{2\pi}{3}\, \sin\dfrac{5\pi}{9} =$ (A) $\dfrac{3}{4}$ (B) $\dfrac{3}{8}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{3}{32}$

    ›Reveal solutionSolution

    Reduce each angle to its acute supplement, then apply the classic identity $\sin20^\circ\sin40^\circ\sin80^\circ=\sqrt3/8$ to get $\tfrac{3}{16}$.

    Concept and Intuition

    Angles like $\dfrac{8\pi}{9}$ (=160°) and $\dfrac{7\pi}{9}$ (=140°) are obtuse, but $\sin(\pi-\theta)=\sin\theta$ folds them down to acute angles that fit the well-known product identity $\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\tfrac14\sin3\theta$ (with $\theta=20^\circ$).

    Step-by-Step Solution

    1. Convert each obtuse angle using $\sin(\pi-\theta)=\sin\theta$: $\sin\dfrac{8\pi}{9}=\sin\left(\pi-\dfrac{8\pi}{9}\right)=\sin\dfrac{\pi}{9}$ (20°); $\sin\dfrac{7\pi}{9}=\sin\dfrac{2\pi}{9}$ (40°); $\sin\dfrac{5\pi}{9}=\sin\dfrac{4\pi}{9}$ (80°). $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$ (120°, and $\sin120^\circ=\sin60^\circ$).
    2. So the product becomes $\sin20^\circ\cdot\sin40^\circ\cdot\dfrac{\sqrt3}{2}\cdot\sin80^\circ = \dfrac{\sqrt3}{2}\left(\sin20^\circ\sin40^\circ\sin80^\circ\right)$. …

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11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.$\cos 12^\circ \cdot \cos 24^\circ \cdot \cos 36^\circ \cdot \cos 48^\circ \cdot \cos 72^\circ \cdot \cos 84^\circ =$ (A) $\dfrac{1}{32}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{1}{64}$ (D) $\dfrac{1}{128}$

    ›Reveal solutionSolution

    Splitting the six-factor product into $\cos36°\cos72°=1/4$ and a doubling-angle chain $\cos12°\cos24°\cos48°\cos84°=1/16$ gives the overall product $1/64$.

    Concept and Intuition

    Products of cosines at angles related by repeated doubling ($\theta,2\theta,4\theta,\dots$) telescope via $\cos\theta\cos2\theta\cdots\cos2^{n-1}\theta=\dfrac{\sin2^n\theta}{2^n\sin\theta}$. Also, $\cos36°\cos72°=1/4$ is a well-known special value from the golden-ratio pentagon identities.

    Step-by-Step Solution

    1. Split off $\cos36°\cos72°$: this is the classical identity $\cos36°\cos72°=\dfrac14$.
    2. Remaining factors: $\cos12°\cos24°\cos48°\cos84°$.
    3. Group $\cos12°\cos24°\cos48°$ — these are $\theta,2\theta,4\theta$ with $\theta=12°$, so by the doubling identity $\cos12°\cos24°\cos48°=\dfrac{\sin96°}{8\sin12°}$.
    4. Multiply by $\cos84°$: note $\cos84°=\sin6°$ and $\sin96°=\sin(90°+6°)=\cos6°$. So the product becomes $\dfrac{\cos6°\sin6°}{8\sin12°}$.
    5. $\cos6°\sin6°=\tfrac12\sin12°$, so this simplifies to $\dfrac{\frac12\sin12°}{8\sin12°}=\dfrac1{16}$. …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.$\cos6^{\circ}\sin24^{\circ}\cos72^{\circ}=$ (A) $-\dfrac{1}{8}$ (B) $-\dfrac{1}{4}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{1}{4}$

    ›Reveal solutionSolution

    This is a product of three trig values at special angles that evaluates to a clean constant, $1/8$.

    Concept and Intuition

    $\cos72°=\sin18°$ and this product of three cosine/sine values at angles related by factors tied to $18°/36°$ is a disguised instance of the well-known identity $\cos\theta\cos(60°-\theta)\cos(60°+\theta)=\tfrac14\cos3\theta$-family of results; here it is most reliably confirmed by direct exact evaluation.

    Step-by-Step Solution

    1. Rewrite $\sin24°=\cos66°$, so the product is $\cos6°\cos66°\cos72°$.
    2. Use $\cos A\cos B = \tfrac12[\cos(A-B)+\cos(A+B)]$ on $\cos6°\cos66°$: $=\tfrac12[\cos60°+\cos72°]=\tfrac12\left[\tfrac12+\cos72°\right]$.
    3. So the full product is $\tfrac12\left[\tfrac12+\cos72°\right]\cos72° = \tfrac14\cos72° + \tfrac12\cos^272°$.
    4. Using the exact value $\cos72°=\dfrac{\sqrt5-1}{4}$: $\cos^272°=\dfrac{(\sqrt5-1)^2}{16}=\dfrac{6-2\sqrt5}{16}=\dfrac{3-\sqrt5}{8}$. …

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