First slide

Compound angles

Question

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

Moderate

A

$\frac{1}{4}$
B

$\frac{1}{2}$
C

$\frac{1}{\sqrt{2}}$
D

$\frac{1}{2 \sqrt{2}}$

Solution

$\begin{array}{l} = \cos^{3} \frac{π}{8} [4 \cos^{3} \frac{π}{8} - 3 \cos \frac{π}{8}] + \sin^{3} \frac{π}{8} [3 \sin \frac{π}{8} - 4 \sin^{3} \frac{π}{8}] \\ = 4 \cos^{6} \frac{π}{8} - 4 \sin^{6} \frac{π}{8} - 3 \cos^{4} \frac{π}{8} + 3 \sin^{4} \frac{π}{8} \\ = 4 [(\cos^{2} \frac{π}{8} - \sin^{2} \frac{π}{8})] [(\sin^{4} \frac{π}{8} + \cos^{4} \frac{π}{8} + \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8})] - 3 [(\cos^{2} \frac{π}{8} - \sin^{2} \frac{π}{8})] \\ = \cos \frac{π}{4} [4 (1 - \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8}) - 3] = \frac{1}{\sqrt{2}} [1 - \frac{1}{2}] = ∣ \frac{1}{2 \sqrt{2}} \end{array}$

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