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a. $\cos^{2}x=\frac{1+\cos2x}{2}$

b. $\sin^{2}x=\frac{1-\cos2x}{2}$

I know what the half-angle identities are—I learned about them in school. However, what I'm confused about is why exactly are these functions above given?

Also, can someone please explain what a proof of this would look like?

Thanks—any help is greatly appreciated. :)

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  • $\begingroup$ You can write identity a as $\cos^{2} (z/2)=\frac{1+\cos(z)}{2}$, wheree $z=2x$. The $z/2$ on the left hand side is half of the $z$ on the right hand side. Thus the name "half-angle identity" $\endgroup$ – Brian Borchers Aug 30 '16 at 23:56
  • $\begingroup$ Usually do people take the square root of both sides? @BrianBorchers $\endgroup$ – 关一骏 Aug 31 '16 at 0:43
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These formulae are a basis for linearising trigonometric polynomials. A 3rd one, missing here, is $$\sin x\cos x=\frac12\sin 2x.$$ They allow integration of trigonometric polynomials.

It's simple to prove if you know the addition formulae: \begin{align*} \cos (x+y)&=\cos x \cos y-\sin x\sin y&\sin (x+y)&=\sin x \cos y+\cos x\sin y\\ \cos (x-y)&=\cos x \cos y+\sin x\sin y&\sin (x-y)&=\sin x \cos y-\cos x\sin y&\\ \end{align*} From these you deduce duplication formulae, setting $y=x$: $$\cos 2x=\cos^2x-\sin^2x,\qquad\sin 2x=2\sin x\cos x,$$ and Pythagoras identity allows to rewrite the first formula as $$\cos 2x=2\cos ^2x-1=1-2\sin^2x.$$ Now the linearisation formulae are just duplication formulae, read backwards.

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$$\cos 2x=\cos { \left( x+x \right) =\cos { x\cos { x-\sin { x\sin { x } =\cos ^{ 2 }{ x-\sin ^{ 2 }{ x } } =\\=\cos ^{ 2 }{ x-\left( 1-\cos ^{ 2 }{ x } \right) =2\cos ^{ 2 }{ x-1 } } } } } } \\ $$ $$\quad \quad \quad \quad \quad \quad \Downarrow \\ \cos ^{ 2 } x=\frac { 1+\cos 2x }{ 2 } $$

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First,$$\cos(x) = \frac{\mathrm{e}^{ix} + \mathrm{e}^{-ix}}{2}$$ \begin{align} \cos^{2}(x) & = \frac{\mathrm{e}^{i2x}+2+\mathrm{e}^{-i2x}}{4} \\ & = \frac{1}{2} + \frac{1}{2} \frac{\mathrm{e}^{i2x} + \mathrm{e}^{-i2x}}{2} \\ & = \frac{1+\cos(2x)}{2} \end{align}

Then,$$\sin(x) = \frac{\mathrm{e}^{ix} - \mathrm{e}^{-ix}}{i2}$$ \begin{align} \sin^{2}(x) & = \frac{\mathrm{e}^{i2x}-2+\mathrm{e}^{-i2x}}{-4} \\ & = \frac{1}{2} - \frac{1}{2} \frac{\mathrm{e}^{i2x} + \mathrm{e}^{-i2x}}{2} \\ & = \frac{1-\cos(2x)}{2} \end{align}

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  • $\begingroup$ Sorry, I'm a bit confused. What are all the e's? $\endgroup$ – 关一骏 Aug 31 '16 at 0:44
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    $\begingroup$ @BirlantisEscheatvc e is called Euler's number, It is everywhere. Check on this site, there are many good question about it. $\endgroup$ – A---B Aug 31 '16 at 1:05
  • $\begingroup$ Cool! Thanks! @A---B $\endgroup$ – 关一骏 Aug 31 '16 at 2:27

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