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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?.

2 Answers 2

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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 $\cos^4y-(-\sin y)^4=(\cos^2y-\sin^2y)(\cos^2y+\sin^2y)=\cos2y$

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    How is $\frac {\pi}{8}+\frac {7\pi}{8}=\frac {\pi}{2}$? – 2017-02-07
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    @tRiGoNoMeTrY, Sorry fro the typo, NeWtoN – 2017-02-07
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    @ lab bhattacharjee, Its Ok. Could you please help me with this one. 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 .... – 2017-02-07
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Hint: Use $cos^2 a=\dfrac{1+cos2a}{2}$ $$\dfrac {4(\dfrac{1+cos2(\dfrac {\pi}{8})}{2})^2 +4(\dfrac{1+cos2(\dfrac {3\pi}{8})}{2})^2}{(\dfrac{1+cos2(\dfrac {7\pi}{8})}{2})^2 - (\dfrac{1+cos2(\dfrac {11\pi}{8})}{2})^2 }=\\ $$ $$\dfrac {4(\dfrac{1+cos\dfrac {\pi}{4}}{2})^2 +4(\dfrac{1+cos\dfrac {3\pi}{4}}{2})^2}{(\dfrac{1+cos\dfrac {7\pi}{4}}{2})^2 - (\dfrac{1+cos\dfrac {11\pi}{4}}{2})^2 }=\\ $$ $$\dfrac {4(\dfrac{1+\dfrac {\sqrt{2}}{2}}{2})^2 +4(\dfrac{1+\dfrac {-\sqrt{2}}{2}}{2})^2}{(\dfrac{1+\dfrac {\sqrt{2}}{2}}{2})^2 - (\dfrac{1+\dfrac {-\sqrt{2}}{2}}{2})^2 }=\\ $$ $$0,\dfrac{\pi}{4},\dfrac{2\pi}{4},\dfrac{3\pi}{4},\dfrac{4\pi}{4}=\pi\\,\dfrac{5\pi}{4},\dfrac{6\pi}{4},\dfrac{7\pi}{4},\dfrac{8\pi}{4}=2\pi\\ \dfrac{9\pi}{4},\dfrac{10\pi}{4},\dfrac{11\pi}{4},\dfrac{12\pi}{4}=3\pi,\\cos(\dfrac{\pi}{4})=\dfrac{\sqrt2}{2},cos(\dfrac{3\pi}{4})=\dfrac{-\sqrt2}{2}\\cos(\dfrac{7\pi}{4})=\dfrac{\sqrt2}{2},cos(\dfrac{11\pi}{4})=\dfrac{-\sqrt2}{2}$$enter image description here

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    Could you please elaborate a bit more/ – 2017-02-07
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    Sure ,but which part ? – 2017-02-07
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    From the part where you plugged in all the numerical values? – 2017-02-07