Answer
The required value of $6{{\sin }^{4}}x$ is $\frac{9}{4}-3\cos 2x+\frac{3}{4}\cos 4x$.
Work Step by Step
We have to find the value of $6{{\sin }^{4}}x$; various sine formulas are used that represent the answer in the form of cos.
$\begin{align}
& 6{{\sin }^{4}}x=6{{\left( \frac{1-\cos 2x}{2} \right)}^{2}} \\
& =6\left( \frac{1-2\cos 2x+{{\cos }^{2}}2x}{4} \right) \\
& =\frac{6-12\cos 2x+6{{\cos }^{2}}2x}{4} \\
& =\frac{3}{2}-3\cos 2x+\frac{3}{2}{{\cos }^{2}}2x
\end{align}$
I
And in step 2, the identity of ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ is used.
$\begin{align}
& 6{{\sin }^{4}}x=\frac{3}{2}-3\cos 2x+\frac{3}{2}\left( \frac{1+\cos 4x}{2} \right) \\
& =\frac{3}{2}-3\cos 2x+\frac{3}{2}\left( \frac{1}{2}+\frac{\cos 4x}{2} \right) \\
& =\frac{3}{2}-3\cos 2x+\frac{3}{4}+\frac{3}{4}\cos 4x \\
& =\frac{9}{4}-3\cos 2x+\frac{3}{4}\cos 4x
\end{align}$
Or, in the last step, $\frac{3}{4}$ is distributed and added to $\frac{3}{2}$.
