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I was to simplify an expression, I came to this:

$$X=2\cos^210^\circ -2\cos^220^\circ$$

But, I couldn't simplify it further.

I checked the answer and it was $\sin10^\circ$. But I can't understand why.

Can anyone help?


EDIT: $\theta$ was $10^\circ$. I myself generalized it.

The two sides are not equivalent. So $10^\circ$ seems to have a special characteristic which we should use here.

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    $\begingroup$ Check $\theta =\frac{\pi}{4}$. $\endgroup$ – Rohan Dec 30 '16 at 11:18
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    $\begingroup$ There's a mistake somewhere ; LHS is even, RHS is odd. What was the expression to simplify? $\endgroup$ – D. Thomine Dec 30 '16 at 11:19
  • $\begingroup$ @Fib1123 It was 10. $\endgroup$ – AHB Dec 30 '16 at 11:23
  • $\begingroup$ @D.Thomine Yes. It was an equation. Not an Identity. Edited. $\endgroup$ – AHB Dec 30 '16 at 11:33
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We can write our equation as: \begin{align} 2\cos^210^\circ -2\cos^2 20^\circ &= 2(\cos 10^\circ + \cos 20^\circ)(\cos 10^\circ -\cos 20^\circ)\\ &=2(2\cos 15^\circ \cos 5^\circ)(2\sin 15^\circ \sin 5^\circ)\\ &=8\sin 15^\circ \cos 15^\circ \sin 5^\circ \cos 5^\circ\\ &= 2(2\sin 15^\circ \cos 15^\circ)(2\sin 5^\circ \cos 5^\circ)\\ &= 2\sin 30^\circ \sin 10^\circ\\ &= \sin 10^\circ. \end{align} Hope it helps.

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Use the “sum to product formulas” \begin{align} \cos10^\circ+\cos20^\circ&=2\cos15^\circ\cos5^\circ\\ \cos10^\circ-\cos20^\circ&=2\sin15^\circ\sin5^\circ \end{align} so your expression is $$ 8\sin15^\circ\cos15^\circ\sin5^\circ\cos5^\circ= 2\sin30^\circ\sin10^\circ=\sin10^\circ $$ In general, $$ 2(\cos^2\theta-\cos^22\theta)= 8\sin\frac{3\theta}{2}\cos\frac{3\theta}{2} \sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin3\theta\sin\theta $$

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\begin{align} X &=2 \cos^210^\circ-1-(2 \cos^220^\circ-1)\\ &= \cos 20^\circ-\cos40^\circ\\ &=2 \sin30^\circ \sin10^\circ\\ &=2\cdot\tfrac12\cdot \sin10^\circ. \end{align}

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