Question

The period of \(\sin^4x+\cos^4x\) is

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi^2}{2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{2}\)

Hint:

If \(\sin(kx)\) has period \(\frac{2\pi}{k}\), then \(\sin^2(kx)\) has period \(\frac{\pi}{k}\).

Answer 1

The Correct Option is D

Step 1: Simplify expression.
\[ \sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x \]
\[ =1-2\sin^2x\cos^2x \]
Step 2: Use identity \(\sin^2x\cos^2x=\frac{1}{4}\sin^22x\).
\[ \sin^4x+\cos^4x=1-2\cdot\frac{1}{4}\sin^22x \]
\[ =1-\frac{1}{2}\sin^22x \]
Step 3: Determine period.
\(\sin^22x\) has period \(\pi/2\) because \(\sin 2x\) has period \(\pi\), and squaring halves it:
\[ T=\frac{\pi}{2} \]
Final Answer:
\[ \boxed{\frac{\pi}{2}} \]