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$2\sin^2(80^\circ) - \sin(70^\circ) = 1$

I can verify with a calculator that equality does hold,
however how do I prove it with trigonometric identities?

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  • $\begingroup$ I would try writing $80 = 75+5$, $70 = 75 - 5$, then using the expressions for $\sin(x+y)$ and $\sin(x-y)$. Remember that you can compute $\sin(75) = \sin(45+30)$ if you need to, using the same strategy. All in degrees, clearly. $\endgroup$ – Ivo Terek Jun 6 '15 at 21:44
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Hint: $\sin 80^{\circ} = \cos 10^{\circ}$ and $\sin 70^{\circ} = \cos 20^{\circ}$; now use a double-angle identity.

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Starting from, $$LHS=2 \sin^280^o-\sin 70^o$$ $$=2 \sin^2(90^o-10^o)-\sin (90^o-20^o)$$ $$=2 \cos^210^o-\cos 20^o$$ $$=2 \cos^210^o-(2 \cos^210^o-1)$$ $$=2 \cos^210^o-2 \cos^210^o+1$$$$=1=RHS$$

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