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I have solved this problem, but I had to use a calculator, how do I solve it without using a calculator(wont be allowed to use a calculator in the examination).

Here is my attempt:

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2 Answers 2

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$$a^2+b^2+c^2=4R^2(\sin^2\frac{\pi}{7}+\sin^2\frac{2\pi}{7}+\sin^2\frac{4\pi}{7})=$$ $$=2R^2\left(3-\cos\frac{2\pi}{7}-\cos\frac{4\pi}{7}-\cos\frac{8\pi}{7}\right)=2R^2\left(3+\frac{1}{2}\right)=7R^2$$ Because $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{8\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{9\pi}{7}-\sin\frac{7\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}$$

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    Actually $\cos\dfrac{8\pi}7=\cos\dfrac{6\pi}7$ as $\dfrac{6\pi}7+\dfrac{8\pi}7=2\pi$ so that angles becomes in Arithmetic progression. See http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – 2017-02-11
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    @lab bhattacharjee Thank you, but all this is very very known. – 2017-02-11
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    This explains why multiplication by $\sin\dfrac\pi7$ works – 2017-02-11
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    @lab bhattacharjee Do you want an interesting problem in trigonometry? – 2017-02-11
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    @Micheal, Why not and why one? – 2017-02-12
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    @lab bhattacharjee I think one it's enough: Let $a$, $b$ and $c$ be roots of the equation $x^3+15x^2-198x+1=0$. Prove that $\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$ – 2017-02-12
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    @lab bhattacharjee The solution you can made from http://math.stackexchange.com/questions/2141207 I think do you understand, why it's trigonometry? :) – 2017-02-12
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HINT:

For $A+B+C=\pi,$ we can prove, $$\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$$

Now $$\cos x\cos2x\cos4x=\dfrac{\sin2x\cos2x\cos4x}{2\sin x}=\dfrac{\sin8x}{8\sin x}$$

$8x=7x+x$

Here $7x=\pi\implies\sin8x=-\sin x$