Question

the value of 16 sin cube theta + 8 cos square theta is​

Answer 1

Answer:

16 \sin^3 \theta+ 8\cos^3 \theta =

To prove:

Simplify.

Solution:

Formula:

\bold{\sin 3\theta= 3\sin \theta -4\sin^3 \theta}\\

\bold{4\sin^3 \theta= 3\sin \theta -\sin 3\theta}\\

\bold{ \cos3\theta= 4\cos^3 \theta-3\cos \theta}

\bold{ 4\cos^3 \theta=\cos3\theta +3\cos \theta}

\Rightarrow 16 \sin^3 \theta+ 8\cos^3 \theta = 4 \cdot 4 \sin^3 \theta+ 2 \cdot 4\cos^3 \theta

= 4 (3\sin \theta- \sin 3\theta)+ 2 (\cos 3\theta+3\cos \theta)\\\\= 12\sin \theta- 4\sin 3\theta+ 2\cos 3\theta+6\cos \theta

The final answer is= 12\sin \theta- 4\sin 3\theta+ 2\cos 3\theta+6\cos \theta \\\\