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How does one derive the following two identities: $\begin{align*} \cos 2\theta &= 1-2\sin^2\theta\\ \sin 2\theta &= 2\sin\theta\cos\theta \end{align*}$

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    @MTurgeon Just a typo.2012-04-01

3 Answers 3

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Hints:

For the $\cos 2\theta$ formula, use the sum identity (with $x=y=\theta$) $ \cos(x+y)=\cos x\cos y - \sin x\sin y, $ followed by the Pythagorean identity $\cos^2 x=1-\sin^2 x $.


For the $\sin 2\theta$ formula , use the sum identity (with $x=y=\theta$) $ \sin(x+y)=\sin x\cos y +\sin y\cos x. $




Or, for the $\cos2\theta$ formula and $0<\theta<\pi/2$, consider the diagram:

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We have $ \cos\theta ={ {1+\cos2\theta}\over\sqrt{2+2\cos 2\theta}}; $ whence $ \sqrt 2\cos\theta=\sqrt{1+1\cos\theta}, $ or, $ 2\cos^2\theta= 1+1\cos2\theta $ From this, we have $ 2-2\sin^2\theta=1+\cos2\theta, $ or $ \cos2\theta =1-2\sin^2\theta. $

(Having this in hand, we could also use the diagram to derive the formula for $\sin2\theta$.)

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    @David Mitra Can you give me the multiple angle formula? I haven't found them in any website.2018-04-01
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$e^{i\theta}$ means the point with angle $\theta$ on the unit circle. Euler formula tells us that $e^{i\theta}=\cos \theta+i\sin \theta$. Then $e^{i2\theta}=\cos^2 \theta-\sin^2 \theta+2i\sin \theta\cos\theta$. Also, we know $e^{i(2\theta)}=\cos 2\theta+i\sin 2\theta$. Compare the real and imaginary parts, and you get the desired result.

All I assume is that you know Euler formula.

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    This is the easiest way to remember/derive it, but obviously one needs to know what complex numbers are.2012-04-01
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The first one follows from:

$\cos 2\theta = \cos(\theta +\theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta .$ Now use the fact $\cos^2 \theta + \sin^2 \theta =1.$

The second one follows from $ \sin 2\theta = \sin(\theta +\theta) = \sin \theta \cos \theta + \cos \theta \sin \theta. $