Why is $1+\cos(\theta)=2\cos^2(\frac{\theta}{2})$
Why is $1+\cos(\theta)=2\cos^2(\frac{\theta}{2})$?
Where this comes from? I don't get it.
From $\sin^2\theta+\cos^2\theta=1$?? I search everything but I really don't find that.
$\endgroup$7 Answers
$\begingroup$$$\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1$$ Then put $$x = 2 \theta$$.
$\endgroup$ $\begingroup$Hint: Use the Law of Cosines on an isosceles triangle whose equal sides are $1$ to find $\cos\theta$. Then use this result to derive. $$1+\cos\theta=2\cos^2(\theta/2)$$
$\endgroup$ $\begingroup$Two basic formulas of trigonometry: \begin{gather} \sin^2\alpha+\cos^2\alpha=1\\[6px] \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \end{gather}
In the second formula, set $\alpha=\beta=\theta/2$: then \begin{align} \cos\theta&= \cos\frac{\theta}{2}\cos\frac{\theta}{2}- \sin\frac{\theta}{2}\sin\frac{\theta}{2}\\[6px] &= \cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\\[6px] &= \cos^2\frac{\theta}{2}-1+\cos^2\frac{\theta}{2}\\[6px] &= 2\cos^2\frac{\theta}{2}-1 \end{align}
$\endgroup$ $\begingroup$Addition formula: $\cos(2z)=\cos^2 z - \sin^2 z = \cos^2 -(1-\cos^2 z) = 2 \cos^2 z - 1.$ Now use $z=\theta /2$
$\endgroup$ $\begingroup$HINT
Use: $\displaystyle \quad \cos \theta=\cos \left (2\cdot\frac{\theta}{2}\right)=\cdots$
$\endgroup$ $\begingroup$Let $f(z)=1+\cos z-2 \cos^2 \dfrac{z}{2}$. Then $$ f'(z)=-\sin z+2 \cos \dfrac{z}{2}\sin \dfrac{z}{2}=-\sin z + \sin z=0. $$ Thus $f(z)$ is a constant for any $z.$ For example $f(z)=f(0)=1+1-2=0.$ $f(z)=0$ follows that $$1+\cos z=2 \cos^2 \dfrac{z}{2}.$$
$\endgroup$ $\begingroup$By definition ($i$ is the complex unit), $$ \cos \theta =\frac{e^{i \theta}+e^{-i \theta}}{2}$$ Now expand the RHS: $$ 2\cos^2 \frac{\theta}{2}=2\left(\frac{e^{i \theta/2}+e^{-i \theta/2}}{2}\right)^2=2\cdot\frac{e^{i \theta}+2+e^{-i \theta}}{4}=1+\frac{e^{i\theta}+e^{-i\theta}}{2}=1+\cos\theta$$
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