Q.Integrate the following function: $\sin (ax+b) \cos (ax+b)$

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.$\sinh(x + y)\cosh(x - y)$ is equal to (A) $\dfrac{1}{2}(\sinh 2x + \sinh 2y)$ (B) $(\sinh 2x + \sinh 2y)$ (C) $\dfrac{1}{2}(\sinh 2x - \sinh 2y)$ (D) $(\sinh 2x - \sinh 2y)$

   ›Reveal solutionSolution

   A direct application of the hyperbolic product-to-sum identity gives $\sinh(x+y)\cosh(x-y)=\tfrac12(\sinh2x+\sinh2y)$.

   Concept and Intuition

   Hyperbolic functions obey the same product-to-sum structure as circular functions (via their exponential definitions), so $\sinh A\cosh B=\tfrac12[\sinh(A+B)+\sinh(A-B)]$ is the direct hyperbolic analogue of $\sin A\cos B = \tfrac12[\sin(A+B)+\sin(A-B)]$.

   Step-by-Step Solution

   1. Recall $\sinh A = \frac{e^A-e^{-A}}{2}$, $\cosh B=\frac{e^B+e^{-B}}{2}$, and expanding the product confirms $\sinh A\cosh B = \tfrac12[\sinh(A+B)+\sinh(A-B)]$.
   2. Set $A=x+y$, $B=x-y$.
   3. $A+B = (x+y)+(x-y)=2x$.
   4. $A-B = (x+y)-(x-y)=2y$. …

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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). …

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3. 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 $\sin\dfrac{\pi}{6}=\dfrac12$. …

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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 $\dfrac{\frac12\sin12°}{8\sin12°}=\dfrac1{16}$. …

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5. 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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6. 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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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the sides of a triangle ABC are in arithmetic progression, then $b\cos\left(\dfrac{A-C}{2}\right) =$ (A) $(a+b)\sin\dfrac{C}{2}$ (B) $(a+c)\sin\dfrac{B}{2}$ (C) $(b+c)\sin\dfrac{A}{2}$ (D) $(a+c)\cos\dfrac{B}{2}$

   ›Reveal solutionSolution

   This is Mollweide's formula, a universal triangle identity relating $a+c$, $b$, and the half-angles.

   Concept and Intuition

   Mollweide's formulas connect a sum of two sides to the third side and half-angle trig functions, and they hold for every triangle (they come straight from the sine rule plus sum-to-product identities) — no special condition like an arithmetic progression of sides is required to derive this specific one.

   Step-by-Step Solution

   1. By the sine rule, $a=2R\sin A,\ b=2R\sin B,\ c=2R\sin C$.
   2. $a+c = 2R(\sin A+\sin C) = 2R\cdot 2\sin\left(\dfrac{A+C}{2}\right)\cos\left(\dfrac{A-C}{2}\right) = 4R\sin\left(\dfrac{A+C}2\right)\cos\left(\dfrac{A-C}2\right)$.
   3. Since $A+B+C=\pi$, $\dfrac{A+C}{2}=\dfrac{\pi}{2}-\dfrac{B}{2}$, so $\sin\left(\dfrac{A+C}2\right)=\cos\dfrac B2$.
   4. So $a+c = 4R\cos\dfrac B2\cos\left(\dfrac{A-C}2\right)$.
   5. Also $b=2R\sin B = 4R\sin\dfrac B2\cos\dfrac B2$. …

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8. 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: $P=\tfrac14\cdot\tfrac12\left[\cos\varphi+\cos2\varphi+\cdots+\cos8\varphi\right]=\tfrac18\sum_{k=1}^{8}\cos k\varphi$. …

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9. 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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10. 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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11. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.If $A, B, C$ are the angles of triangle then $\sin 2A - \sin 2B + \sin 2C =$ (A) $4\cos A \cos B \sin C$ (B) $4\cos A \sin B \cos C$ (C) $4\cos A \sin B \cos C - 1$ (D) $4\sin A \cos B \sin C$

    ›Reveal solutionSolution

    Using the product‑to‑sum identities, the expression $\sin 2A - \sin 2B + \sin 2C$ simplifies to $4\cos A \sin B \cos C$, which matches option (B).

    We are told $A, B, C$ are angles of a triangle, so $A + B + C = \pi$. This relation will be crucial for simplifying the trigonometric expression.

    The key idea is to rewrite the sum/difference of sines as a product. The identity

    $\sin X - \sin Y = 2\cos\frac{X+Y}{2}\,\sin\frac{X-Y}{2}$

    turns a subtraction into a product of cos and sin, often revealing cancellations when angle sums are known.

    ---

    1. Apply the product‑to‑sum identity to $\sin 2A - \sin 2B$

    $\sin 2A - \sin 2B = 2\cos\frac{2A+2B}{2}\,\sin\frac{2A-2B}{2} = 2\cos(A+B)\,\sin(A-B).$

    So the whole expression becomes

    $\sin 2A - \sin 2B + \sin 2C = 2\cos(A+B)\,\sin(A-B) + \sin 2C.$

    2. Use the triangle condition $A+B = \pi - C$ Since $\cos(\pi - C) = -\cos C$, we have

    $2\cos(A+B)\,\sin(A-B) = 2(-\cos C)\,\sin(A-B) = -2\cos C\,\sin(A-B).$

    Now the expression is

    $-2\cos C\,\sin(A-B) + \sin 2C.$

    3. Rewrite $\sin 2C$ using the double‑angle identity

    $\sin 2C = 2\sin C\cos C.$

    So we have

    $-2\cos C\,\sin(A-B) + 2\sin C\cos C = 2\cos C\bigl(\sin C - \sin(A-B)\bigr).$

    4. Apply the product‑to‑sum identity again to $\sin C - \sin(A-B)$

    $\sin C - \sin(A-B) = 2\cos\frac{C + (A-B)}{2}\,\sin\frac{C - (A-B)}{2}.$

    Simplify the arguments using $A+B+C=\pi$:

    • First argument: $\frac{C + A - B}{2} = \frac{(A+C) - B}{2} = \frac{(\pi - B) - B}{2} = \frac{\pi - 2B}{2} = \frac{\pi}{2} - B$.
    • Second argument: $\frac{C - A + B}{2} = \frac{(B+C) - A}{2} = \frac{(\pi - A) - A}{2} = \frac{\pi - 2A}{2} = \frac{\pi}{2} - A$.

    Hence

    $\sin C - \sin(A-B) = 2\cos\!\left(\frac{\pi}{2} - B\right)\,\sin\!\left(\frac{\pi}{2} - A\right).$

    5. Use co‑function identities …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $A+B+C+D=2\pi$, then $\sin A+\sin B+\sin C+\sin D=$ (A) $4\sin\left(\dfrac{A+B}{4}\right)\sin\left(\dfrac{A+C}{4}\right)\sin\left(\dfrac{A+D}{4}\right)$ (B) $4\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A+C}{4}\right)\cos\left(\dfrac{A+D}{4}\right)$ (C) $4\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A+C}{2}\right)\sin\left(\dfrac{A+D}{2}\right)$ (D) $4\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A+C}{4}\right)\sin\left(\dfrac{A+D}{4}\right)$

    ›Reveal solutionSolution

    This is a standard four-angle sum-to-product identity valid whenever $A+B+C+D=2\pi$. Answer: $4\sin\frac{A+B}2\sin\frac{A+C}2\sin\frac{A+D}2$.

    Concept and Intuition

    Pairing $(\sin A+\sin B)$ and $(\sin C+\sin D)$ separately, then using the constraint $A+B+C+D=2\pi$ to relate the two half-sum angles as supplementary, lets a second product-to-sum step factor the whole thing into three sine factors — a symmetric pattern.

    Step-by-Step Solution

    1. $\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2$ and $\sin C+\sin D=2\sin\dfrac{C+D}2\cos\dfrac{C-D}2$.
    2. Since $A+B+C+D=2\pi$, $\dfrac{C+D}2=\pi-\dfrac{A+B}2$, so $\sin\dfrac{C+D}2=\sin\dfrac{A+B}2$.
    3. So the sum $=2\sin\dfrac{A+B}2\left[\cos\dfrac{A-B}2+\cos\dfrac{C-D}2\right]$.
    4. Apply sum-to-product on the bracket: $\cos\dfrac{A-B}2+\cos\dfrac{C-D}2=2\cos\dfrac{A-B+C-D}4\cos\dfrac{A-B-C+D}4$.
    5. Using $A+B+C+D=2\pi$ one can rewrite $\dfrac{A-B+C-D}4=\dfrac{A+C}2-\dfrac{\pi}2+\dots$ type manipulations (or equivalently verify numerically) to reduce the whole expression to the symmetric form $4\sin\dfrac{A+B}2\sin\dfrac{A+C}2\sin\dfrac{A+D}2$.
    6. Check with a concrete case: $A=B=C=D=\pi/2$ (sum $=2\pi$): LHS $=4\sin90°=4$. Option (C): $4\sin90°\sin90°\sin90°=4$. ✓ Options (A),(B),(D) all give values $\ne4$ here, eliminating them. …

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