16 sin cube theta+8 cos cube theta=
Answers
Answer:
Given:
16 \sin^3 \theta+ 8\cos^3 \theta =16sin
3
θ+8cos
3
θ=
To prove:
Simplify.
Solution:
Formula:
\begin{gathered}\bold{\sin 3\theta= 3\sin \theta -4\sin^3 \theta}\\\end{gathered}
sin3θ=3sinθ−4sin
3
θ
\begin{gathered}\bold{4\sin^3 \theta= 3\sin \theta -\sin 3\theta}\\\end{gathered}
4sin
3
θ=3sinθ−sin3θ
\bold{ \cos3\theta= 4\cos^3 \theta-3\cos \theta}cos3θ=4cos
3
θ−3cosθ
\bold{ 4\cos^3 \theta=\cos3\theta +3\cos \theta}4cos
3
θ=cos3θ+3cosθ
\Rightarrow 16 \sin^3 \theta+ 8\cos^3 \theta = 4 \cdot 4 \sin^3 \theta+ 2 \cdot 4\cos^3 \theta⇒16sin
3
θ+8cos
3
θ=4⋅4sin
3
θ+2⋅4cos
3
θ
\begin{gathered}= 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\end{gathered}
=4(3sinθ−sin3θ)+2(cos3θ+3cosθ)
=12sinθ−4sin3θ+2cos3θ+6cosθ
The final answer is= \begin{gathered}12\sin \theta- 4\sin 3\theta+ 2\cos 3\theta+6\cos \theta \\\\\end{gathered}
12sinθ−4sin3θ+2cos3θ+6cosθ
Given:
16 \sin^3 \theta+ 8\cos^3 \theta =
To prove:
Simplify.
Solution:
Formula:
{\sin 3\theta= 3\sin \theta -4\sin^3 \theta}\\
{4\sin^3 \theta= 3\sin \theta -\sin 3\theta}\\
{ \cos3\theta= 4\cos^3 \theta-3\cos \theta}
{ 4\cos^3 \theta=\cos3\theta +3\cos \theta}
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 \\= 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 \
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