Q.Integrate the following function: $\cos 2x \cos 4x \cos 6x$

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$\cos\dfrac{6\pi}{17} \cos\dfrac{10\pi}{17} \cos\dfrac{12\pi}{17} \cos\dfrac{14\pi}{17} =$ (A) $-\dfrac{1}{16}$ (B) $\dfrac{1}{16}$ (C) $-16$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Tests the classical "17-gon" cosine-product trick (product-to-sum + the identity $\cos k\varphi=\cos(17-k)\varphi$ where $\varphi=2\pi/17$). The value is $-\dfrac1{16}$.

   Concept and Intuition

   The angles $6\pi/17,10\pi/17,12\pi/17,14\pi/17$ are all integer multiples of $\varphi=\dfrac{2\pi}{17}$: they equal $3\varphi,5\varphi,6\varphi,7\varphi$. Products of cosines at rational multiples of $\pi$ over a prime denominator collapse beautifully once you repeatedly apply $\cos A\cos B=\tfrac12[\cos(A-B)+\cos(A+B)]$, together with the periodicity fact $\cos(k\varphi)=\cos((17-k)\varphi)$ (because $17\varphi=2\pi$).

   Step-by-Step Solution

   1. Let $P=\cos3\varphi\cos5\varphi\cos6\varphi\cos7\varphi$, $\varphi=2\pi/17$.
   2. Pair up: $\cos3\varphi\cos5\varphi=\tfrac12[\cos2\varphi+\cos8\varphi]$ and $\cos6\varphi\cos7\varphi=\tfrac12[\cos\varphi+\cos13\varphi]$. Since $\cos13\varphi=\cos4\varphi$ (as $17-13=4$), this is $\tfrac12[\cos\varphi+\cos4\varphi]$.
   3. So $P=\tfrac14[\cos2\varphi+\cos8\varphi][\cos\varphi+\cos4\varphi]$. Expand into four products and reduce each with product-to-sum, using $\cos9\varphi=\cos8\varphi$ and $\cos12\varphi=\cos5\varphi$:
      • $\cos2\varphi\cos\varphi=\tfrac12[\cos\varphi+\cos3\varphi]$
      • $\cos2\varphi\cos4\varphi=\tfrac12[\cos2\varphi+\cos6\varphi]$
      • $\cos8\varphi\cos\varphi=\tfrac12[\cos7\varphi+\cos8\varphi]$
      • $\cos8\varphi\cos4\varphi=\tfrac12[\cos4\varphi+\cos5\varphi]$
   4. Adding: $P=\tfrac14\cdot\tfrac12\left[\cos\varphi+\cos2\varphi+\cdots+\cos8\varphi\right]=\tfrac18\sum_{k=1}^{8}\cos k\varphi$.
   5. Since $\sum_{k=1}^{16}\cos k\varphi=-1$ (all 17th roots of unity sum to 0) and $\cos k\varphi=\cos(17-k)\varphi$ pairs $k=1..8$ with $k=9..16$, we get $2\sum_{k=1}^{8}\cos k\varphi=-1$, so $\sum_{k=1}^{8}\cos k\varphi=-\tfrac12$.
   6. Hence $P=\tfrac18\times\left(-\tfrac12\right)=-\dfrac1{16}$. (Numeric check: $\cos63.53^\circ\cos105.88^\circ\cos127.06^\circ\cos148.24^\circ\approx-0.0625$.)

   Common Mistakes

   • Trying to evaluate each cosine individually instead of using the symmetry $\cos k\varphi=\cos(17-k)\varphi$.
   • Forgetting that the full root-of-unity sum over $k=1..16$ is $-1$, not $0$.

   ✓Final answer

   The correct option is (A) — $-\dfrac{1}{16}$.

   ANSWER: A

2. 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 $\cos72°=\dfrac{\sqrt5-1}{4}$: $\cos^272°=\dfrac{(\sqrt5-1)^2}{16}=\dfrac{6-2\sqrt5}{16}=\dfrac{3-\sqrt5}{8}$.
   5. Substitute: $\tfrac14\cdot\dfrac{\sqrt5-1}{4} + \tfrac12\cdot\dfrac{3-\sqrt5}{8} = \dfrac{\sqrt5-1}{16}+\dfrac{3-\sqrt5}{16} = \dfrac{2}{16}=\dfrac18$.

   Common Mistakes

   • Attempting to force-fit the $\cos\theta\cos(60-\theta)\cos(60+\theta)$ pattern directly without first converting $\sin24°$ to a cosine form, which doesn't line up cleanly.
   • Sign/arithmetic slips simplifying $(\sqrt5-1)^2$.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{8}$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$\cos\dfrac{\pi}{7} \cos\dfrac{2\pi}{7} \cos\dfrac{3\pi}{7} \cos\dfrac{4\pi}{7} \cos\dfrac{5\pi}{7} \cos\dfrac{6\pi}{7} =$ (A) $\dfrac{1}{8}$ (B) $-\dfrac{1}{16}$ (C) $\dfrac{1}{32}$ (D) $-\dfrac{1}{64}$

   ›Reveal solutionSolution

   Fold the six-term product down to a square of the classic identity $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}=\frac18$, picking up a minus sign from the supplementary-angle pairing. Answer: $-\frac{1}{64}$.

   Concept and Intuition

   Angles like $\frac{4\pi}{7},\frac{5\pi}{7},\frac{6\pi}{7}$ are each supplementary to $\frac{3\pi}{7},\frac{2\pi}{7},\frac{\pi}{7}$ respectively (they sum to $\pi$), and $\cos(\pi-x)=-\cos x$. This symmetry lets the six-term product collapse into (a sign) times the square of the well-known three-term product $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}$, whose value ($1/8$) is a standard result provable via the $\sin(2^n x)$ doubling trick.

   Step-by-Step Solution

   1. Note the supplementary pairs (each pair sums to $\pi$): $\frac{4\pi}{7}=\pi-\frac{3\pi}{7}$, $\frac{5\pi}{7}=\pi-\frac{2\pi}{7}$, $\frac{6\pi}{7}=\pi-\frac{\pi}{7}$.
   2. So $\cos\frac{4\pi}{7}=-\cos\frac{3\pi}{7}$, $\cos\frac{5\pi}{7}=-\cos\frac{2\pi}{7}$, $\cos\frac{6\pi}{7}=-\cos\frac{\pi}{7}$.
   3. The full product becomes:

   $\left[\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{3\pi}{7}\right]\cdot\left[(-\cos\tfrac{3\pi}{7})(-\cos\tfrac{2\pi}{7})(-\cos\tfrac{\pi}{7})\right] = -\left[\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{3\pi}{7}\right]^2$

   (the three minus signs from the second bracket multiply to an overall $-1$).

   4. Use the standard identity $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}=\frac18$ (derivable by multiplying and dividing by $\sin\frac{\pi}{7}$ and repeatedly applying $2\sin x\cos x=\sin 2x$, which telescopes down to $\sin\frac{8\pi}{7}$ related back to $\sin\frac{\pi}{7}$).

   5. So the six-term product $= -\left(\frac18\right)^2 = -\frac{1}{64}$.

   Common Mistakes

   • Forgetting the sign flip that comes from the three supplementary-angle substitutions (an odd number of sign flips, so the overall sign is negative) — a very easy place to drop a minus sign.
   • Misremembering the three-term identity's value (it's $\frac18$, not $\frac14$ or $\frac{1}{16}$).

   ✓Final answer

   The correct option is (D) — $-\dfrac{1}{64}$.

   ANSWER: D

4. 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 $\dfrac{\frac12\sin12°}{8\sin12°}=\dfrac1{16}$.
   6. Total product $=\dfrac14\times\dfrac1{16}=\dfrac1{64}$.

   Common Mistakes

   • Trying to apply the doubling-angle telescoping identity to all six factors at once, when $36°,72°$ don't fit the $\theta,2\theta,4\theta,\dots$ chain starting from $12°$ (since $12\times2\times2\times2=96\ne36$) — they must be handled as a separate golden-ratio identity.
   • Sign/complement slip converting $\sin96°$ to $\cos6°$.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{64}$.

   ANSWER: C

5. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.$\sin 21^\circ \cos 9^\circ - \cos 84^\circ \cos 6^\circ =$ (A) 1 (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Convert $\cos84^\circ$ to $\sin6^\circ$ and expand both products using sum-to-product identities; the $\sin12^\circ$ terms cancel exactly, leaving $\dfrac14$.

   Concept and Intuition

   When an expression mixes cosines and sines of complementary-looking angles ($84^\circ=90^\circ-6^\circ$), converting everything to a common trig function often reveals hidden cancellation via product-to-sum formulas.

   Step-by-Step Solution

   1. Note $\cos84^\circ = \sin(90^\circ-84^\circ)=\sin6^\circ$.
   2. So $\cos84^\circ\cos6^\circ = \sin6^\circ\cos6^\circ = \dfrac12\sin12^\circ$ (double-angle identity).
   3. For the first term, use $\sin X\cos Y = \dfrac12[\sin(X+Y)+\sin(X-Y)]$ with $X=21^\circ,Y=9^\circ$: $\sin21^\circ\cos9^\circ = \dfrac12[\sin30^\circ+\sin12^\circ] = \dfrac12\left[\dfrac12+\sin12^\circ\right] = \dfrac14+\dfrac12\sin12^\circ$.
   4. Subtract: $\left(\dfrac14+\dfrac12\sin12^\circ\right) - \dfrac12\sin12^\circ = \dfrac14$.

   Common Mistakes

   • Not recognizing $\cos84^\circ=\sin6^\circ$, which is the key simplification that makes the $\sin12^\circ$ terms cancel.
   • Sign errors in the product-to-sum expansion.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{4}$.

   ANSWER: B

6. 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 $\sin20^\circ\cdot\sin40^\circ\cdot\dfrac{\sqrt3}{2}\cdot\sin80^\circ = \dfrac{\sqrt3}{2}\left(\sin20^\circ\sin40^\circ\sin80^\circ\right)$.
   3. Use the identity $\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac14\sin3\theta$ with $\theta=20^\circ$: note $40^\circ=60^\circ-20^\circ$ and $80^\circ=60^\circ+20^\circ$, so $\sin20^\circ\sin40^\circ\sin80^\circ = \dfrac14\sin60^\circ = \dfrac14\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{8}$.
   4. Combine: total $=\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{8} = \dfrac{3}{16}$.

   Common Mistakes

   • Forgetting to fold the obtuse angles $8\pi/9$ and $7\pi/9$ back to their acute equivalents before applying the identity.
   • Misremembering the triple-angle product identity's coefficient (it's $\tfrac14$, not $\tfrac12$).

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{16}$.

   ANSWER: C

7. 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 $n=8$: $\displaystyle\prod_{k=1}^{7}\cos\left(\dfrac{k\pi}{16}\right) = \dfrac{\sqrt8}{2^7} = \dfrac{2\sqrt2}{128}=\dfrac{\sqrt2}{64}$.
   4. Squaring: $\displaystyle\prod_{k=1}^{7}\cos^2\left(\dfrac{k\pi}{16}\right) = \left(\dfrac{\sqrt2}{64}\right)^2 = \dfrac{2}{4096}=\dfrac{1}{2048}$.
   5. Total product $= 2^7\times\dfrac{1}{2048} = \dfrac{128}{2048}=\dfrac{1}{16}$.

   Common Mistakes

   • Using the wrong analog formula (the sine-product identity $\prod\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}$ is different from the cosine one used here) — mixing them up gives a wrong constant.
   • Forgetting the factor $2^7$ that comes from converting each of the 7 terms via the half-angle identity.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{16}$.

   ANSWER: A

8. 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: $f''(x)=\dfrac12(-2\sin2x)+(-4\sin4x)-\dfrac32(-6\sin6x)=-\sin2x-4\sin4x+9\sin6x$.
   7. Matching to $a\sin bx+c\sin dx+e\sin kx$: $a=-1,b=2$; $c=-4,d=4$; $e=9,k=6$.
   8. $(a+c+e)-(b+d+k)=(-1-4+9)-(2+4+6)=4-12=-8$.

   Common Mistakes

   • Sign errors when repeatedly differentiating $\sin(nx)$ twice (each differentiation of $\sin$ gives $\cos$, then $\cos$ gives $-\sin$, so two derivatives bring back $-n^2\sin(nx)$ — but here we only need the first derivative's coefficient pattern before differentiating again, so it's easy to drop a sign along the way).
   • Forgetting the constant $\frac14$ factor carried through both differentiations.

   ✓Final answer

   The correct option is (B) — $-8$.

   ANSWER: B

9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The number of solutions of the equation $4\cos 2\theta \cos 3\theta = \sec \theta$ in the interval $[0, 2\pi]$ is (A) $12$ (B) $8$ (C) $16$ (D) $4$

   ›Reveal solutionSolution

   Clearing $\sec\theta$ and expanding the triple product of cosines with sum-to-product identities collapses the equation to $\cos2x(2\cos x+1)=0$ with $x=2\theta$; carefully counting all solutions of both factors over the full range gives $12$ solutions.

   Concept and Intuition

   Equations mixing products of cosines at different multiples of $\theta$ ($2\theta,3\theta$) with $\sec\theta$ are best handled by first clearing the secant (multiplying by $\cos\theta$, keeping in mind $\cos\theta\ne0$ is required for $\sec\theta$ to exist), then repeatedly applying product-to-sum and double-angle identities to reduce everything to a sum of cosines that can be factored.

   Step-by-Step Solution

   1. $4\cos2\theta\cos3\theta=\sec\theta\Rightarrow 4\cos\theta\cos2\theta\cos3\theta=1$ (multiplying by $\cos\theta\ne0$).
   2. $2\cos\theta\cos3\theta=\cos2\theta+\cos4\theta$ (product-to-sum), so $4\cos\theta\cos2\theta\cos3\theta=2\cos2\theta(\cos2\theta+\cos4\theta)=2\cos^22\theta+2\cos2\theta\cos4\theta$.
   3. $2\cos^22\theta=1+\cos4\theta$ and $2\cos2\theta\cos4\theta=\cos2\theta+\cos6\theta$, so the LHS becomes $1+\cos4\theta+\cos2\theta+\cos6\theta$.
   4. Equation: $1+\cos2\theta+\cos4\theta+\cos6\theta=1\Rightarrow \cos2\theta+\cos4\theta+\cos6\theta=0$.
   5. Let $x=2\theta\in[0,4\pi]$: $\cos x+\cos2x+\cos3x=0$. Using $\cos x+\cos3x=2\cos2x\cos x$: $2\cos2x\cos x+\cos2x=0\Rightarrow \cos2x(2\cos x+1)=0$.
   6. Branch $\cos2x=0$: $x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$. For $x\in[0,4\pi)$, $k=0,\dots,7$ gives 8 solutions.
   7. Branch $\cos x=-\tfrac12$: $x=\dfrac{2\pi}{3}+2k\pi$ or $x=\dfrac{4\pi}{3}+2k\pi$. For $x\in[0,4\pi)$: $x=\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{8\pi}{3},\tfrac{10\pi}{3}$ — 4 solutions.
   8. None of these $x$-values correspond to $\cos\theta=0$ (i.e. $x=\pi$ or $3\pi$), so all 12 values of $x$ give valid $\theta=x/2\in[0,2\pi]$.
   9. Total number of solutions $=8+4=12$.

   Common Mistakes

   • Forgetting to check that $\cos\theta\ne0$ throughout (since we multiplied by $\cos\theta$ to clear $\sec\theta$) — here it turns out none of the 12 solutions violate this, but the check must still be done.
   • Losing solutions by not tracking the full range $x\in[0,4\pi)$ correctly when converting back from $x=2\theta$.

   ✓Final answer

   The correct option is (A) — $12$.

   ANSWER: A

10. 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 $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{4\pi}{7}=-\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=-\dfrac18$.
    4. Exact values: $\cos36^\circ=\dfrac{1+\sqrt5}{4}\cdot ...$ numerically $\cos\dfrac{\pi}{5}=\cos36^\circ=\dfrac{\sqrt5+1}{4}\times2^{?}$; using the standard result $\cos36^\circ\cos72^\circ=\dfrac14$ (which follows from $\cos36^\circ=\dfrac{\sqrt5+1}{4}$-type exact forms), we get $\cos\dfrac{\pi}{5}\cos\dfrac{2\pi}{5}=\dfrac14$.
    5. Multiply everything: $4\cos\dfrac{\pi}{7}\cos\dfrac{\pi}{5}\cos\dfrac{2\pi}{7}\cos\dfrac{2\pi}{5}\cos\dfrac{4\pi}{7}=4\times\left(-\dfrac18\right)\times\dfrac14=-\dfrac18$.

    Common Mistakes

    • Using $\cos\frac{3\pi}{7}$ in place of $\cos\frac{4\pi}{7}$ directly without accounting for the sign flip $\cos\frac{4\pi}{7}=-\cos\frac{3\pi}{7}$.
    • Misremembering the exact value/product for $\cos36^\circ\cos72^\circ$.

    ✓Final answer

    The correct option is (A) — $-\dfrac{1}{8}$.

    ANSWER: A

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. $=\frac12\sin\frac{\pi}{8}\cdot\frac12\sin\frac{3\pi}{8} = \frac14\sin\frac{\pi}{8}\sin\frac{3\pi}{8}$.
    5. Since $\sin\frac{3\pi}{8}=\cos\frac{\pi}{8}$, this is $\frac14\sin\frac{\pi}{8}\cos\frac{\pi}{8}=\frac18\sin\frac{\pi}{4}=\frac18\cdot\frac{\sqrt2}{2}=\frac{\sqrt2}{16}$.

    Common Mistakes

    • Missing that $\sin\frac{5\pi}{16}$ and $\sin\frac{7\pi}{16}$ are cosines of different smaller angles, not the same one.

    ✓Final answer

    The correct option is (A) — $\frac{\sqrt{2}}{16}$.

    ANSWER: A

12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If $A+B+C=\dfrac{\pi}{4}$, then $\sin 4A+\sin 4B+\sin 4C=$ (A) $4\cos 2A\cos 2B\cos 2C$ (B) $4\sin 2A\sin 2B\sin 2C$ (C) $1+4\sin 2A\sin 2B\sin 2C$ (D) $1+4\cos 2A\cos 2B\cos 2C$

    ›Reveal solutionSolution

    This tests recognizing a disguised "triangle-angle" identity after a substitution. Answer: $\sin4A+\sin4B+\sin4C=4\cos2A\cos2B\cos2C$.

    Concept and Intuition

    The classic identity "if $X+Y+Z=\pi$ then $\sin X+\sin Y+\sin Z=4\cos\frac X2\cos\frac Y2\cos\frac Z2$" is usually stated for a triangle's angles, but it's really just a trigonometric identity that holds for any three angles summing to $\pi$ — it doesn't care where the angles came from. Here $A+B+C=\pi/4$ looks unrelated at first, but multiplying every angle by 4 turns the condition into exactly $X+Y+Z=\pi$ with $X=4A$ etc., unlocking the identity immediately.

    Step-by-Step Solution

    1. Given $A+B+C=\dfrac{\pi}{4}$. Multiply by 4: $4A+4B+4C=\pi$.
    2. Let $X=4A,\ Y=4B,\ Z=4C$, so $X+Y+Z=\pi$.
    3. Apply the standard identity (valid whenever three angles sum to $\pi$): $\sin X+\sin Y+\sin Z=4\cos\dfrac X2\cos\dfrac Y2\cos\dfrac Z2$.
    4. Here $\dfrac X2=2A,\ \dfrac Y2=2B,\ \dfrac Z2=2C$.
    5. So $\sin4A+\sin4B+\sin4C=4\cos2A\cos2B\cos2C$.

    Common Mistakes

    • Trying to expand $\sin4A,\sin4B,\sin4C$ individually via multiple-angle formulas and grinding through algebra — much slower and error-prone than spotting the disguised $X+Y+Z=\pi$ pattern.
    • Misremembering the identity as involving $\sin\frac X2\sin\frac Y2\sin\frac Z2$ instead of cosines (that alternate form applies to a different triangle identity, for $\cos X+\cos Y+\cos Z$).

    ✓Final answer

    The correct option is (A) — $4\cos 2A\cos 2B\cos 2C$.

    ANSWER: A