Question

If $f\left(x\right)={\cos }^{2}x+{\cos }^2(\frac{\pi }{3}+x)-\cos x\cos (\frac{\pi }{3}+x)$ then $4f\left(\frac{\pi }{8}\right)$ is equal to

Answer 1

$f\left(x\right)={\cos }^{2}x+{\cos }^2(\frac{\pi }{3}+x)-\cos x\cos (\frac{\pi }{3}+x)\cos (a+b)=\cos a\cos b-\sin a\sin bf\left(x\right)={\cos }^{2}x+{(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x)}^2-\cos x(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x)f\left(x\right)={\cos }^{2}x+\frac{{\cos }^{2}x}{4}+\frac{3{\sin }^{2}x}{4}-2.\frac{1}{2}\cos x.\frac{\sqrt{3}}{2}\sin x-\frac{{\cos }^{2}x}{2}+\cos x.\frac{\sqrt{3}}{2}\sin xf\left(x\right)=\frac{3}{4}({\cos }^{2}x+{\sin }^{2}x)f\left(x\right)=\frac{3}{4}4f\left(\frac{\pi }{8}\right)=3$