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

$\cos42^\circ\cos78^\circ=\tfrac12(\cos120^\circ+\cos36^\circ)=\tfrac12(-\tfrac12+\cos36^\circ)$.

Multiplying by $\cos18^\circ$ and evaluating numerically: …

Concept and Intuition

Sum-to-product turns sin x + sin y and cos x + cos y into forms sharing the common factor 2 cos((x-y)/2), so their ratio is simply tan of the half-sum (x+y)/2.

Step-by-Step Solution

1. a = sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2).
2. b = cos x + cos y = 2 cos((x+y)/2) cos((x-y)/2).
3. a/b = tan((x+y)/2) = tan(pi/3) = sqrt3. …

Concept and Intuition

Using the sum-to-product identity cosP+cosQ, the bracket collapses to 2cos x cos y; halving it exactly cancels the product term cos x cos y.

Step-by-Step Solution

1. $\cos(x+y) = \cos x\cos y - \sin x\sin y$.
2. $\cos(y-x) = \cos y\cos x + \sin y\sin x$.
3. Sum $= 2\cos x\cos y$, so $\tfrac{1}{2}[\cdots] = \cos x\cos y$. …

$\sin7x+\sin5x=2\sin6x\cos x,\qquad \cos7x+\cos5x=2\cos6x\cos x.$ …

Concept and Intuition

The double-angle identity sin A cos A = (1/2) sin 2A, plus the co-function relation sin(pi/2 - x) = cos x, collapses the expression onto cos(pi/14).

Step-by-Step Solution

1. sin(3pi/14) cos(3pi/14) = (1/2) sin(6pi/14) = (1/2) sin(3pi/7).
2. 3pi/7 = pi/2 - pi/14, so sin(3pi/7) = cos(pi/14). …

[!TLDR]

With $\cos36^\circ\cos72^\circ=\tfrac14$, the expression $4\cos36^\circ\cos72^\circ=1$.

Concept

The standard exact values are $\cos36^\circ=\dfrac{\sqrt5+1}{4}$ and $\cos72^\circ=\dfrac{\sqrt5-1}{4}$ (they can also be derived from multiple-angle identities in the NCERT/CBSE trigonometry syllabus).

Solution …