Q.Integrate the following function: $\sin x \sin 2x \sin 3x$

Showing the 12 most recent of 40 on this concept.

1. 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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2. 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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3. 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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4. 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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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\cos 13^\circ \sin 17^\circ \sin 21^\circ \cos 47^\circ =$ (A) $\dfrac{1}{32}(1+\sqrt{2}-\sqrt{3})$ (B) $\dfrac{1}{16}(1+\sqrt{3}+\sqrt{5})$ (C) $\dfrac{1}{16}(2+\sqrt{3}-\sqrt{5})$ (D) $\dfrac{1}{32}(1+2\sqrt{3}-\sqrt{5})$

   ›Reveal solutionSolution

   Evaluating $\cos13^\circ\sin17^\circ\sin21^\circ\cos47^\circ$ numerically and matching against the four surd expressions identifies the value as $\dfrac1{32}(1+2\sqrt3-\sqrt5)$.

   Concept and Intuition

   Products of several trig ratios at "odd" degree angles are usually meant to be collapsed via repeated product-to-sum identities down to a combination of surds ($\sqrt2,\sqrt3,\sqrt5$ typically arise from $15^\circ,18^\circ,36^\circ$-type angles hiding inside sums/differences of the given angles). When the algebra gets heavy, a fast and reliable check is to evaluate the product numerically to several decimal places and test which of the printed surd expressions reproduces that decimal value exactly.

   Step-by-Step Solution

   1. Compute each factor: $\cos13^\circ\approx0.974370$, $\sin17^\circ\approx0.292372$, $\sin21^\circ\approx0.358368$, $\cos47^\circ\approx0.681998$.
   2. Multiply progressively: $0.974370\times0.292372\approx0.284878$; $\times0.358368\approx0.102091$; $\times0.681998\approx0.069626$.
   3. Evaluate option (D): $1+2\sqrt3-\sqrt5=1+3.46410-2.23607=2.22803$; divide by $32$: $2.22803/32=0.0696259$. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.$\sin\dfrac{\pi}{12}\sin\dfrac{2\pi}{12}\sin\dfrac{3\pi}{12}\sin\dfrac{4\pi}{12}\sin\dfrac{5\pi}{12}\sin\dfrac{6\pi}{12}=$ (A) $\dfrac{\sqrt3}{16\sqrt2}$ (B) $\dfrac{\sqrt3}{8\sqrt2}$ (C) $\dfrac{1}{32}$ (D) $\dfrac{1}{16}$

   ›Reveal solutionSolution

   A product of six sines at multiples of $15°$ collapses using the complementary pair $\sin\theta\sin(90°-\theta)=\sin\theta\cos\theta$. Answer: $\dfrac{\sqrt3}{16\sqrt2}$.

   Concept and Intuition

   The six angles $\pi/12,\dots,6\pi/12$ are $15°,30°,45°,60°,75°,90°$. Two of them, $15°$ and $75°$, are complementary ($15°+75°=90°$), so their product simplifies via $\sin\theta\cos\theta=\tfrac12\sin2\theta$. The remaining four are standard angles whose sines are memorized values.

   Step-by-Step Solution

   1. Write the product as $\sin15°\cdot\sin30°\cdot\sin45°\cdot\sin60°\cdot\sin75°\cdot\sin90°$.
   2. Pair $\sin15°\sin75°=\sin15°\cos15°=\tfrac12\sin30°=\tfrac12\cdot\tfrac12=\tfrac14$.
   3. The remaining factors are $\sin30°=\tfrac12$, $\sin45°=\tfrac{\sqrt2}{2}$, $\sin60°=\tfrac{\sqrt3}{2}$, $\sin90°=1$.
   4. Multiply everything: $\tfrac14\times\tfrac12\times\tfrac{\sqrt2}{2}\times\tfrac{\sqrt3}{2}\times1=\dfrac{\sqrt6}{32}$. …

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7. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\cos^3 x \sin 4x = \sum_{r=0}^{n} a_r \sin rx$ $\forall x \in \mathbb{R}$, then $a_3+a_5 : a_1+a_7 =$ (A) $1:3$ (B) $1:1$ (C) $2:1$ (D) $3:1$

   ›Reveal solutionSolution

   Expanding $\cos^3x\sin4x$ as a sum of sines via product-to-sum identities gives coefficients $a_1=a_7=\frac18$, $a_3=a_5=\frac38$, so the requested ratio is $3:1$.

   Concept and Intuition

   $\cos^3x$ reduces to a linear combination of $\cos x$ and $\cos3x$ (a standard multiple-angle reduction), turning the product into a sum of two simpler $\cos\cdot\sin$ products, each of which splits into two sine terms via product-to-sum formulas.

   Step-by-Step Solution

   1. $\cos^3x=\dfrac{3\cos x+\cos3x}{4}$ (from $\cos3x=4\cos^3x-3\cos x$).
   2. $\cos^3x\sin4x=\dfrac34\cos x\sin4x+\dfrac14\cos3x\sin4x$.
   3. $\cos x\sin4x=\dfrac12[\sin(4x+x)+\sin(4x-x)]=\dfrac12(\sin5x+\sin3x)$.
   4. $\cos3x\sin4x=\dfrac12[\sin(4x+3x)+\sin(4x-3x)]=\dfrac12(\sin7x+\sin x)$.
   5. Combine: $\cos^3x\sin4x=\dfrac34\cdot\dfrac12(\sin5x+\sin3x)+\dfrac14\cdot\dfrac12(\sin7x+\sin x)=\dfrac38\sin5x+\dfrac38\sin3x+\dfrac18\sin7x+\dfrac18\sin x$. …

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8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If the general solution set of $\sin x + 3\sin 3x + \sin 5x = 0$ is S, then $\{\sin \alpha \mid \alpha \in S\} =$ (A) $\{1, -1, 0\}$ (B) $\left\{\dfrac{1}{2}, \dfrac{-1}{2}, 0, 1, -1\right\}$ (C) $\left\{\dfrac{\sqrt{3}}{2}, 0, \dfrac{-\sqrt{3}}{2}\right\}$ (D) $\left\{1, -1, \dfrac{\sqrt{3}}{2}, 0, \dfrac{-\sqrt{3}}{2}\right\}$

   ›Reveal solutionSolution

   This tests sum-to-product factoring of a trig equation; the solution set of $x$-values reduces to $x=n\pi/3$, and $\{\sin\alpha\}$ over that set is $\{\frac{\sqrt3}{2},0,-\frac{\sqrt3}{2}\}$.

   Concept and Intuition

   $\sin x+\sin5x$ combines via sum-to-product into a factor of $\sin3x$, which is already present in the middle term, so the whole equation factors neatly — a common trick in trigonometric equations with terms in arithmetic-progression multiples of $x$.

   Step-by-Step Solution

   1. $\sin x+\sin5x=2\sin\left(\dfrac{x+5x}{2}\right)\cos\left(\dfrac{5x-x}{2}\right)=2\sin3x\cos2x$.
   2. The equation becomes $2\sin3x\cos2x+3\sin3x=0\Rightarrow \sin3x(2\cos2x+3)=0$.
   3. $2\cos2x+3=0\Rightarrow\cos2x=-\dfrac32$, impossible since $\cos2x\in[-1,1]$.
   4. So $\sin3x=0\Rightarrow 3x=n\pi\Rightarrow x=\dfrac{n\pi}{3},\ n\in\mathbb Z$. This is $S$. …

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9. 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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10. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.In a triangle ABC, $\sin 2A + \sin 2B + \sin 2C =$ (A) 4 sinA sinB sinC (B) 2 sinA sinB sinC (C) 4 cosA cosB cosC (D) 2 sinA cosB cosC

    ›Reveal solutionSolution

    A classic triangle-angle-sum identity: the sum of double-angle sines equals four times the product of the sines.

    Concept and Intuition

    Because $A+B+C=\pi$ in any triangle, sum-to-product manipulations on $\sin2A+\sin2B+\sin2C$ collapse neatly into a product form involving all three angles.

    Step-by-Step Solution

    1. $\sin2A+\sin2B = 2\sin(A+B)\cos(A-B)$.
    2. Since $A+B+C=\pi$, $A+B=\pi-C$, so $\sin(A+B)=\sin C$.
    3. So $\sin2A+\sin2B=2\sin C\cos(A-B)$.
    4. Also $\sin2C=2\sin C\cos C = 2\sin C\cos(\pi-(A+B)) = -2\sin C\cos(A+B)$.
    5. Sum: $\sin2A+\sin2B+\sin2C = 2\sin C[\cos(A-B)-\cos(A+B)]$.
    6. Using $\cos(A-B)-\cos(A+B)=2\sin A\sin B$: total $=2\sin C\cdot2\sin A\sin B=4\sin A\sin B\sin C$. …

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11. 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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12. 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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