Question

16 sin cube theta+8 cos cube theta =​

Answer 1

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

Answer 2

Answer:

16sin

3

θ+8cos

3

θ=

To prove:

Simplify.

Solution:

Formula:

\begin{lgathered}\bold{\sin 3\theta= 3\sin \theta -4\sin^3 \theta}\\\end{lgathered}

sin3θ=3sinθ−4sin

3

θ

\begin{lgathered}\bold{4\sin^3 \theta= 3\sin \theta -\sin 3\theta}\\\end{lgathered}

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{lgathered}= 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{lgathered}

=4(3sinθ−sin3θ)+2(cos3θ+3cosθ)

=12sinθ−4sin3θ+2cos3θ+6cosθ

The final answer is= \begin{lgathered}12\sin \theta- 4\sin 3\theta+ 2\cos 3\theta+6\cos \theta \\\\\end{lgathered}

12sinθ−4sin3θ+2cos3θ+6cosθ