Question:
The value of
\[
\cos^4 x
\]
is
The value of
\[
\cos^4 x
\]
is
Show Hint
To reduce higher powers of trigonometric functions, repeatedly use:
\[
\cos^2x=\frac{1+\cos 2x}{2}
\]
and
\[
\sin^2x=\frac{1-\cos 2x}{2}
\]
These are called power reduction formulas.
Updated On: Jun 25, 2026
- \(\dfrac{3}{8}+\dfrac{1}{2}\cos 2x+\dfrac{1}{8}\cos 4x\)
- \(\dfrac{3}{8}-\dfrac{1}{2}\cos 2x+\dfrac{1}{8}\cos 4x\)
- \(\dfrac{3}{8}-\dfrac{1}{8}\cos 4x+\dfrac{1}{2}\cos 2x\)
- \(\dfrac{1}{8}\cos 4x+\dfrac{1}{2}\cos 2x-\dfrac{3}{8}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Express \(\cos^4x\) as \((\cos^2x)^2\).
We have \[ \cos^4x=(\cos^2x)^2 \] Using the identity \[ \cos^2x=\frac{1+\cos 2x}{2}, \] we get \[ \cos^4x=\left(\frac{1+\cos 2x}{2}\right)^2 \] \[ =\frac{1}{4}(1+\cos 2x)^2 \]
Step 2: Expand the square.
\[ \cos^4x = \frac{1}{4}\left(1+2\cos 2x+\cos^22x\right) \] Now use \[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \] with \[ \theta=2x \] Then, \[ \cos^22x=\frac{1+\cos 4x}{2} \] Substitute this value: \[ \cos^4x = \frac{1}{4}\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right) \]
Step 3: Simplify the expression.
\[ \cos^4x = \frac{1}{4}\left(\frac{2+4\cos 2x+1+\cos 4x}{2}\right) \] \[ = \frac{1}{8}\left(3+4\cos 2x+\cos 4x\right) \] Distributing, \[ \cos^4x = \frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x \]
Step 4: Final conclusion.
Therefore, \[ \boxed{ \cos^4x= \frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x } \]
We have \[ \cos^4x=(\cos^2x)^2 \] Using the identity \[ \cos^2x=\frac{1+\cos 2x}{2}, \] we get \[ \cos^4x=\left(\frac{1+\cos 2x}{2}\right)^2 \] \[ =\frac{1}{4}(1+\cos 2x)^2 \]
Step 2: Expand the square.
\[ \cos^4x = \frac{1}{4}\left(1+2\cos 2x+\cos^22x\right) \] Now use \[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \] with \[ \theta=2x \] Then, \[ \cos^22x=\frac{1+\cos 4x}{2} \] Substitute this value: \[ \cos^4x = \frac{1}{4}\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right) \]
Step 3: Simplify the expression.
\[ \cos^4x = \frac{1}{4}\left(\frac{2+4\cos 2x+1+\cos 4x}{2}\right) \] \[ = \frac{1}{8}\left(3+4\cos 2x+\cos 4x\right) \] Distributing, \[ \cos^4x = \frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x \]
Step 4: Final conclusion.
Therefore, \[ \boxed{ \cos^4x= \frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x } \]
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