The value of $\cos ^4 x$ is
The value of $\cos ^4 x$ is
- $\frac{3}{8}+\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x$
- $\frac{3}{8}-\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x$
- $\frac{3}{8}-\frac{1}{8} \cos 4 x+\frac{1}{2} \cos 2 x$
- $\frac{1}{8} \cos 4 x+\frac{1}{2} \cos 2 x-\frac{3}{8}$
Solution
Given expression is $\cos ^4 x$.
$
\begin{aligned}
& \cos 2 x=2 \cos ^2 x-1 \\
& \cos ^2 x=\frac{1+\cos 2 x}{2}
\end{aligned}
$
Take, $\cos ^4 \mathrm{x}=\left(\cos ^2 \mathrm{x}\right)^2=\left(\frac{1+\cos 2 \mathrm{x}}{2}\right)^2$
$
\begin{aligned}
& =\frac{1+\cos ^2 2 x+2 \cos 2 x}{4} \\
& =\frac{1}{4}+\left(\frac{1+\cos 4 x}{2}\right) \times \frac{1}{4}+\frac{\cos x}{2} \\
& =\frac{1}{4}+\frac{1}{8}+\frac{\cos 4 x}{8}+\frac{\cos 2 x}{2} \\
& =\frac{3}{8}+\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
Practice more Trigonometric Ratios & Identities questions on Aicharya