AP EAMCET · Maths · Trigonometric Ratios & Identities
The value of \(\cos ^4 x\) is
- A \(\frac{3}{8}+\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x\)
- B \(\frac{3}{8}-\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x\)
- C \(\frac{3}{8}-\frac{1}{8} \cos 4 x+\frac{1}{2} \cos 2 x\)
- D \(\frac{1}{8} \cos 4 x+\frac{1}{2} \cos 2 x-\frac{3}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{8}+\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x\)
Step-by-step Solution
Detailed explanation
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, \(\cos ^4 \mathrm{x}=\left(\cos ^2 \mathrm{x}\right)^2=\left(\frac{1+\cos 2 \mathrm{x}}{2}\right)^2\)…
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