Q.Integrate the following function: $\sin 3x \cos 4x$

Showing the 12 most recent of 40 on this concept.

1. 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

2. 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

3. 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

4. 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: $\cos^3x\sin4x=\dfrac34\cdot\dfrac12(\sin5x+\sin3x)+\dfrac14\cdot\dfrac12(\sin7x+\sin x)=\dfrac38\sin5x+\dfrac38\sin3x+\dfrac18\sin7x+\dfrac18\sin x$.
   6. Reading off: $a_1=\tfrac18,\ a_3=\tfrac38,\ a_5=\tfrac38,\ a_7=\tfrac18$ (others zero).
   7. $(a_3+a_5):(a_1+a_7)=\left(\tfrac38+\tfrac38\right):\left(\tfrac18+\tfrac18\right)=\tfrac34:\tfrac14=3:1$.

   Common Mistakes

   • Using the wrong triple-angle reduction ($\cos3x=4\cos^3x-3\cos x\Rightarrow\cos^3x=\frac{3\cos x+\cos3x}{4}$ — easy to mix up the sign/coefficient).
   • Errors in the product-to-sum formula's sign convention.

   ✓Final answer

   The correct option is (D) — $3:1$.

   ANSWER: D

5. 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

6. 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

7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The value of $\sin(\frac{5\pi}{24}) \cdot \cos(\frac{\pi}{24})$ is (A) $\frac{1+\sqrt2}{4}$ (B) $1+\sqrt2$ (C) $\frac{1-\sqrt2}{4}$ (D) $1-\sqrt2$

   ›Reveal solutionSolution

   A direct application of the product-to-sum formula turns the awkward angles $5\pi/24$ and $\pi/24$

   into the familiar $\pi/4$ and $\pi/6$; the value is $\dfrac{1+\sqrt2}{4}$.

   Concept and Intuition

   Products of sine and cosine at "ugly" angles often simplify beautifully once you notice that their

   sum and difference are standard angles. Here $\frac{5\pi}{24}+\frac{\pi}{24}=\frac{6\pi}{24}=\frac{\pi}{4}$

   and $\frac{5\pi}{24}-\frac{\pi}{24}=\frac{4\pi}{24}=\frac{\pi}{6}$ — both angles whose sine we know exactly.

   This is exactly the situation the product-to-sum identity is built for.

   Step-by-Step Solution

   1. Recall $\sin A\cos B=\dfrac12\left[\sin(A+B)+\sin(A-B)\right]$.
   2. Here $A=\dfrac{5\pi}{24}$, $B=\dfrac{\pi}{24}$, so $A+B=\dfrac{\pi}{4}$, $A-B=\dfrac{\pi}{6}$.
   3. $\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}$ and $\sin\dfrac{\pi}{6}=\dfrac12$.
   4. So the product $=\dfrac12\left(\dfrac{\sqrt2}{2}+\dfrac12\right)=\dfrac12\cdot\dfrac{\sqrt2+1}{2}=\dfrac{\sqrt2+1}{4}$.

   Common Mistakes

   • Using $\sin A\sin B$ or $\cos A\cos B$ identities instead of the correct $\sin A\cos B$ form.
   • Arithmetic slip turning $\tfrac{\sqrt2+1}{4}$ into $\sqrt2+1$ by forgetting the outer $\tfrac12$ factor twice.

   ✓Final answer

   The correct option is (A) — $\dfrac{1+\sqrt2}{4}$.

   ANSWER: A

8. 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

9. 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

10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.$\displaystyle\sum_{k=0}^{12} \dfrac{1}{\sin\left((k+1)\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)\sin\left(\dfrac{k\pi}{6}+\dfrac{\pi}{4}\right)} =$ (A) $2(\sqrt{3}+1)$ (B) $2(3-\sqrt{3})$ (C) $2(2-\sqrt{3})$ (D) $2(\sqrt{3}-1)$

    ›Reveal solutionSolution

    This is a telescoping trigonometric sum using the cotangent-difference identity. The answer is (D).

    Concept and Intuition

    Whenever consecutive terms in a sum $\dfrac{1}{\sin A_k \sin A_{k-1}}$ have a constant angular gap $d=A_k-A_{k-1}$, the identity $\dfrac{\sin(A-B)}{\sin A\sin B}=\cot B-\cot A$ turns each term into a difference of cotangents, so the whole sum telescopes to just the first and last cotangent.

    Step-by-Step Solution

    1. Let $A_k = k\dfrac{\pi}{6}+\dfrac{\pi}{4}$ for $k=0,\dots,13$. Each term is $\dfrac{1}{\sin A_{k+1}\sin A_k}$, and $A_{k+1}-A_k=\pi/6$ always.
    2. Using $\dfrac{\sin(\pi/6)}{\sin A_{k+1}\sin A_k} = \cot A_k - \cot A_{k+1}$, each term equals $\dfrac{1}{\sin(\pi/6)}(\cot A_k-\cot A_{k+1}) = 2(\cot A_k - \cot A_{k+1})$.
    3. Summing $k=0$ to $12$ telescopes: $\sum = 2(\cot A_0 - \cot A_{13})$.
    4. $A_0=\pi/4$, so $\cot A_0=1$.
    5. $A_{13}=13\pi/6+\pi/4 = 29\pi/12$. Since $\cot$ has period $\pi$, and $29\pi/12-2\pi=5\pi/12$, $\cot(29\pi/12)=\cot(5\pi/12)=\cot75°$.
    6. $\tan75°=2+\sqrt3$, so $\cot75°=\dfrac{1}{2+\sqrt3}=2-\sqrt3$.
    7. Sum $=2[1-(2-\sqrt3)] = 2(\sqrt3-1)$.

    Common Mistakes

    • Forgetting the $\dfrac{1}{\sin(\pi/6)}=2$ scale factor introduced by the identity.
    • Reducing $29\pi/12$ incorrectly (it's more than $2\pi$, needs one full subtraction of $2\pi$ before recognizing the reference angle).

    ✓Final answer

    The correct option is (D) — $2(\sqrt3-1)$.

    ANSWER: D

11. 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. $\tan2x=1$.
    6. $\tan2x=1 \Rightarrow 2x = n\pi+\frac{\pi}{4} \Rightarrow x=\frac{n\pi}{2}+\frac{\pi}{8},\ n\in\mathbb{Z}$.

    Common Mistakes

    • Dividing both sides by the factor $(2\cos x - 3)$ without checking it can never be zero (a valid step here, but one must justify it rather than just assume it).
    • Forgetting the factor of $\frac12$ from $2x$ when converting $2x=n\pi+\frac\pi4$ back to $x$.

    ✓Final answer

    The correct option is (B) — $\dfrac{n\pi}{2}+\dfrac{\pi}{8},\ n\in\mathbb{Z}$.

    ANSWER: B

12. 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