Problem 372: (2002 AMC12P, 17)
Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is $\text{(A) }1-\sqrt{2}\sin{x} \qquad$ $ \text{(B) }-1+\sqrt{2}\cos{x} \qquad$ $ \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad$ $ \text{(D) }\cos{x} - \sin{x} \qquad$ $ \text{(E) }\cos{2x}$
Solution
- $\sin^4x+4\cos^2x=\sin^4x+4(1-\sin^2x)=(\sin^2x-2)^2$, so $\sqrt{\sin^4x+4\cos^2x}=2-\sin^2x=1+\cos^2x$.
- Similarly $\cos^4x+4\sin^2x=(\cos^2x-2)^2$, so $\sqrt{\cos^4x+4\sin^2x}=2-\cos^2x=1+\sin^2x$.
- $f(x)=(1+\cos^2x)-(1+\sin^2x)=\cos^2x-\sin^2x=\cos2x$.
- The answer is (E) $\cos2x$.
For $0<x<\frac\pi4$, let $f(x)=\sqrt{1+\sin2x}-\sqrt{1-\sin2x}$. What is $f\left(\frac\pi{12}\right)$?
