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$Old$ 06-15-2008, 03:51 AM

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$Default$ Trigonometry Identity Rearrangement

Can someone show me how $\sin ^22\theta$ equal to $1-\cos 4\theta$ ?

Thanks in advance.

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$Old$ 06-15-2008, 03:54 AM

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Hello,

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Originally Posted by Air $View Post$

Can someone show me how $\sin ^22\theta$ equal to $1-\cos 4\theta$ ?

Thanks in advance.

$\cos 4 \theta=\cos (2 \cdot 2 \theta)$

Using the identity $\cos 2a=1-2 \sin^2a$ , we get :

$\cos 4 \theta=1-2 \sin^2 2\theta$

$\sin^2 2 \theta=\frac{1-\cos 4 \theta}{\color{red}2}$

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$\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{ and } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}$

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$Old$ 06-15-2008, 03:58 AM

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...So if we had something like:

$\sin^2 6\theta$ , would that be $\frac{1 - \cos 12 \theta}{2}$ ?

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$Old$ 06-15-2008, 03:59 AM

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Quote:

Originally Posted by Air $View Post$

...So if we had something like:

$\sin^2 6\theta$ , would that be $\frac{1 - \cos 12 \theta}{2}$ ?

Yes =)

Always refer to double angle properties ^^
It's just a matter of substitution.

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shinhidora production

$\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{ and } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}$

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