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Let $f(x) = \sin^{2}{x} + \sin^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$ and $g(x) = \left\{\begin{array}{rcc} 2x,\quad0\le x<1 \\x + \frac{1}{4},\quad 1 \le x<2 \end{array}\right.$ , then find $g\{f(x)\}$.

I am really confused on this one.

2 Answers 2

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Note that $f(x)$ can be written as $f(x) = \sin^{2}{x} + \sin^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$ $=1-\cos^{2}{x} + 1-\cos^{2}\left(x + \frac{\pi}{3}\right) + \cos{x}\cos\left(x + \frac{\pi}{3}\right)$ $=2-\frac{3}{4}\cos^{2}{x}-\left[\frac{1}{2}\cos{x}-\cos(x+\frac{\pi}{3})\right]^2.$ Since $\cos(x+\frac{\pi}{3})=\cos x\cos\frac{\pi}{3}-\sin x\sin\frac{\pi}{3}=\frac{1}{2}\cos{x}-\frac{\sqrt{3}}{2}\sin x,$ we have $f(x) = 2-\frac{3}{4}\cos^{2}{x}-\Big(\frac{\sqrt{3}}{2}\sin x\Big)^2=2-\frac{3}{4}\cos^{2}{x}-\frac{3}{4}\sin^{2}{x}=2-\frac{3}{4}=\frac{5}{4}.$ Therefore, $g(f(x))=g(\frac{5}{4})=\frac{5}{4}+\frac{1}{4}=\frac{3}{2}.$

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    Thanks! I thought I got it, but I was wrong. Your solution really helped me.2011-12-11
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At first, you have to realize that there are basically two caes you have to consider. The first one is when $0. Then you compose the functions according to the rule you have written above, i.e., in the intervall for which the condition I have imposed on $f$ holds you write $g(f) = 2\cdot f$. When $1<=f<2$, you have to write: $g(f)=f + \dfrac{1}{4}$. Then you're done.