The value of $\cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ is:

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Trigonometric Ratios and Identities

The value of

cos^{3} (\frac{π}{8}) \cdot cos (\frac{3 π}{8}) + sin^{3} (\frac{π}{8}) \cdot sin (\frac{3 π}{8})

is:

Answer: (D)

Using the triple-angle forms

cos 3 θ = 4 cos^{3} θ - 3 cos θ

and

sin 3 θ = 3 sin θ - 4 sin^{3} θ

with

θ = \frac{π}{8}

, the expression becomes

4 (cos^{6} \frac{π}{8} - sin^{6} \frac{π}{8}) - 3 (cos^{4} \frac{π}{8} - sin^{4} \frac{π}{8})

. Each difference factors through

cos^{2} \frac{π}{8} - sin^{2} \frac{π}{8} = cos \frac{π}{4}

, leaving

cos \frac{π}{4} [4 - 4 cos^{2} \frac{π}{8} sin^{2} \frac{π}{8} - 3] = cos \frac{π}{4} [1 - sin^{2} \frac{π}{4}]

, where

4 cos^{2} \frac{π}{8} sin^{2} \frac{π}{8} = sin^{2} \frac{π}{4} = \frac{1}{2}

. This gives

\frac{1}{2} (1 - \frac{1}{2}) = \frac{1}{2 2}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option