Prove that: $$\frac {4(\cos^4 \frac {\pi}{8} + \cos^4 \frac {3\pi}{8})}{\cos^4 \frac {7\pi}{8} - \cos^4 \frac {11\pi}{8}}=2\sqrt {2} \sin \frac {\pi}{3}.\tan \frac {\pi}{3}$$
My Attempt:
$$L.H.S= \frac {4(\cos^4 \frac {\pi}{8} + \cos^4 \frac {3\pi}{8})}{\cos^4 \frac {7\pi}{8} - \cos^4 \frac {11\pi}{8}}$$ $$=\frac {4((\cos^2 \frac {\pi}{8})^2 + (\cos^2 \frac {3\pi}{8})^2)}{\cos^4 \frac {\pi}{8} - \cos^4 \frac {3\pi}{8}}$$.
What should I do next?.
HINT:
$$\dfrac\pi8+\dfrac{3\pi}8=\dfrac\pi2$$
So, $\cos\dfrac{3\pi}8=\sin\dfrac\pi8$
Now $\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x+\sin^2x=1-\dfrac{(2\sin x\cos x)^2}2=1-\dfrac{\sin^22x}2$
Similarly, $\dfrac{11\pi}8-\dfrac{7\pi}8=\dfrac\pi2$
$\cos\dfrac{11\pi}8=\cos\left(\dfrac\pi2+\dfrac{7\pi}8\right)=-\sin\dfrac{7\pi}8$
Now $\cos^4y-(-\sin y)^4=(\cos^2y-\sin^2y)(\cos^2y+\sin^2y)=\cos2y$
Hint:
Use $cos^2 a=\dfrac{1+cos2a}{2}$
$$\dfrac {4(\dfrac{1+cos2(\dfrac {\pi}{8})}{2})^2 +4(\dfrac{1+cos2(\dfrac {3\pi}{8})}{2})^2}{(\dfrac{1+cos2(\dfrac {7\pi}{8})}{2})^2 - (\dfrac{1+cos2(\dfrac {11\pi}{8})}{2})^2 }=\\
$$
$$\dfrac {4(\dfrac{1+cos\dfrac {\pi}{4}}{2})^2 +4(\dfrac{1+cos\dfrac {3\pi}{4}}{2})^2}{(\dfrac{1+cos\dfrac {7\pi}{4}}{2})^2 - (\dfrac{1+cos\dfrac {11\pi}{4}}{2})^2 }=\\
$$
$$\dfrac {4(\dfrac{1+\dfrac {\sqrt{2}}{2}}{2})^2 +4(\dfrac{1+\dfrac {-\sqrt{2}}{2}}{2})^2}{(\dfrac{1+\dfrac {\sqrt{2}}{2}}{2})^2 - (\dfrac{1+\dfrac {-\sqrt{2}}{2}}{2})^2 }=\\
$$
$$0,\dfrac{\pi}{4},\dfrac{2\pi}{4},\dfrac{3\pi}{4},\dfrac{4\pi}{4}=\pi\\,\dfrac{5\pi}{4},\dfrac{6\pi}{4},\dfrac{7\pi}{4},\dfrac{8\pi}{4}=2\pi\\
\dfrac{9\pi}{4},\dfrac{10\pi}{4},\dfrac{11\pi}{4},\dfrac{12\pi}{4}=3\pi,\\cos(\dfrac{\pi}{4})=\dfrac{\sqrt2}{2},cos(\dfrac{3\pi}{4})=\dfrac{-\sqrt2}{2}\\cos(\dfrac{7\pi}{4})=\dfrac{\sqrt2}{2},cos(\dfrac{11\pi}{4})=\dfrac{-\sqrt2}{2}$$
math.stackexchange.com/questions/2132698/assign-to-each-side-b-of-a-convex-polygon-p-the-maximum-area-of-a-triangle
it would be a favor ....
– pi-π Feb 07 '17 at 09:09