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I need help solving this:

$$ \sin 2x\sin 3x-\cos 2x\cos 3x>\sin 10x. $$

I derived formulas for $\sin 3x$ and $\cos 3x$, but substituting them just gives me the polynomial of fifth degree on LHS. I doubt I have to derive $\sin 10x$, there has to be some interesting shortcut.

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Would the angle-addition formulas help? – Matthew Leingang Feb 4 '13 at 19:29
$\cos (a+b)=\cos a\cos b-\sin a\sin b$ – L. F. Feb 4 '13 at 19:30
Oh gosh, $\cos 5x(2\sin 5x+1)<0$... Well thanks. I guess I need some rest. – Lazar Ljubenović Feb 4 '13 at 19:32
up vote 4 down vote accepted

Note that $$\sin(2x) \sin(3x) - \cos(2x) \cos(3x) = -\cos(5x)$$ and $$\sin(10x) = 2 \sin(5x) \cos(5x)$$ Hence, we need to solve for $$-\cos(5x) > 2 \sin(5x) \cos(5x) \text{ i.e. } \cos(5x)(1+2\sin(5x)) < 0$$ Hence, either $\left(\cos(5x) >0 \text{ and } 1+2\sin(5x) < 0 \right)$ or $\left(\cos(5x) <0 \text{ and } 1+2\sin(5x) > 0 \right)$.

Now you should be able to find the appropriate $x$.

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the right hand side of you first equation should be $-\cos(5x)$ and not $+\cos(5x)$ as you've worked with in the rest of your problem, so you'll need to rework the entire inequality. – amWhy Feb 4 '13 at 19:47
@amWhy Thanks. Updated. – user17762 Feb 4 '13 at 19:51

Hint: use $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b),\;$$ to obtain $$\;\sin 2x \sin3x - \cos(2x)\cos(3x) = - \cos(5x)$$

And use the fact that $$\sin(10x) = 2 \sin(5x) \cos(5x)$$

and solve $$2 \sin(5x) \cos(5x) + \cos 5x < 0$$ $$\implies \cos 5x(2 \sin(5x) + 1) < 0$$

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