Question

The value of \[ \cos^4 x \] is

• A. \(\dfrac{3}{8}+\dfrac{1}{2}\cos 2x+\dfrac{1}{8}\cos 4x\)
• B. \(\dfrac{3}{8}-\dfrac{1}{2}\cos 2x+\dfrac{1}{8}\cos 4x\)
• C. \(\dfrac{3}{8}-\dfrac{1}{8}\cos 4x+\dfrac{1}{2}\cos 2x\)
• D. \(\dfrac{1}{8}\cos 4x+\dfrac{1}{2}\cos 2x-\dfrac{3}{8}\)

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.

Answer 1

The Correct Option is A

Step 1: Write cos^4 x as a square.
$ \cos^4 x = (\cos^2 x)^2 $.
Step 2: Use the power-reduction identity for cos^2 x.
We know $ \cos^2 x = \frac{1 + \cos 2x}{2} $. So: $ \cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{(1 + \cos 2x)^2}{4} $.
Step 3: Expand the square in the numerator.
$(1 + \cos 2x)^2 = 1 + 2\cos 2x + \cos^2 2x $. So $ \cos^4 x = \frac{1 + 2\cos 2x + \cos^2 2x}{4} $.
Step 4: Use the power-reduction identity for cos^2(2x).
Apply $ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} $ with $ \theta = 2x $: $ \cos^2 2x = \frac{1 + \cos 4x}{2} $.
Step 5: Substitute and simplify.
\[ \cos^4 x = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{\frac{2 + 4\cos 2x + 1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8} \]
Step 6: Write out the final form and state the answer.
\[ \cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \] \[ \boxed{\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x} \]