The value of 16 sin cube theta + 8 cos cube theta is equal to
Answers
Answered by
4
Answer:
Given:
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 \\\\
Step-by-step explanation:
Answered by
0
Answer:
yes upper answer is correct dear
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