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solve $\sin 2x = 0.5$

I am looking at the answer to this question and it does not use the double angle formula to make this:

$\sin 2x = 2\sin x \cos x$

And instead use

$2x = \arcsin 0.5$

Why is the double angle formula not used in this situation?

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  • $\begingroup$ don't forget $2x=\pi-arcsin0.5$ and its multiplicities... $\endgroup$ – imranfat Apr 13 '16 at 21:23
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    $\begingroup$ I'll race you. You try to solve $2\sin x\,\cos x = \tfrac12$ and I'll try to solve $2x = \arcsin \tfrac12$. Ready... set... go! $\endgroup$ – David Richerby Apr 14 '16 at 5:03
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Using the double angle formula here ends you up with two things that must multiply together to become $0.5$, and we would need to solve a nasty quadratic to discover $\sin x$ and $\cos x$. Just using $\arcsin$ directly makes your life easier: $\arcsin 0.5$ is very easy to obtain, and we reduce the rest of the problem to simple manipulation.


To give some sense here, using double angle formula you get $$2\sin x \cos x = \frac{1}{2}$$

This with the pythagorean identity $\sin^2 x + \cos^2 x = 1$ can combine to make the following two equations:

$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{3}{2}$$ $$\sin^2 x - 2\sin x \cos x + \cos^2 x = \frac{1}{2}$$ Factor the left side: $$(\sin x + \cos x)^2 = \frac{3}{2}$$ $$(\sin x - \cos x)^2 = \frac{1}{2}$$ Take the square root: $$\sin x + \cos x = \pm\sqrt{\frac{3}{2}}$$ $$\sin x - \cos x = \pm\sqrt{\frac{1}{2}}$$ Add the two equations: $$2\sin x = \pm\sqrt{\frac{3}{2}} \pm \sqrt{\frac{1}{2}}$$ And now finally we can discover the sine of the original angle: $$\sin x = \pm \frac{\sqrt{6}\pm\sqrt{2}}{4}$$ And then we need to apply $\arcsin$ and do some fiddling with "which equivalent angles do I take?". What a mess. Instead, just find $\arcsin \frac{1}{2} = 2x$ and use supplemental and coterminal angles to find the other values of $2x$ that work.

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    $\begingroup$ snazzy approach, i would have just given up trying after getting to $2\sin x \cos x$ $\endgroup$ – MichaelChirico Apr 14 '16 at 0:41

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