I need to find the frequency of the following trigonometric function.$y=\sin^4(x)+\cos^4(x)$ The "answers" section says the answer is: $F_y=\frac{\pi}{2}$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1): $\cos(2x)=1-\sin^2(x)$ $\sin^2(x)=\frac{1-\cos(2x)}{2}$ $\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$ Finding $\cos(2x)^2$ frequency: $\cos(4x)=2\cos^2(2x)-1$ $\cos^2(2x)=\frac{\cos(4x)+1}{2}$ $f_1=\frac{2\pi}{4}=\frac{\pi}{2}$ Finding $\sin(x)^2$ frequency: $\cos(2x)=1-2\sin^2(x)$ $\sin^2(x)=\frac{1-\cos(2x)}{2}$ $f_2=\frac{2\pi}{2}=\pi$ $F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$ $F_1=\frac{\pi}{2}\times2=\pi$ Finding $cos(x)^4$ frequency (I'll call it F2): $\cos(2x)=2\cos^2(x)-1$ $\cos^2(x)=\frac{\cos(2x)+1}{2}$ $\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$ Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1): $f_3=\frac{2\pi}{2}=\pi$ $F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$ $F_2=\frac{\pi}{2}\times2=\pi$ Finding $y$'s frequency: $F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$ $F_y=\pi\times1=\pi$