Trigonometric substitution

Question

Been out of touch with trigonometry for some time now. Need help proving this expression.

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

Any help will be appreciated. Thanks.

Why the down vote? This is outright rude, I post my first question and get a down vote. If you can't help, just move on. — , Apr 17 '14 at 15:24
did you mean $\frac{\sin^2(x)}2 = \frac{1}{2}(1 - \cos^2(x))$ — P.K., Apr 17 '14 at 15:31

score 1 · Answer 1 · edited Apr 18 '14 at 07:38

1

Hint: 1) $\cos(2a)=\cos(a+a)=2\cos^2(a)-1=1-2\sin^2(a)$ (using $\sin^2(a)+\cos^2(a)=1$). Now use the result with $a=x/2$.

edited Apr 18 '14 at 07:38

JRN

6,566

answered Apr 17 '14 at 15:31

voldemort

13,182

score 0 · Accepted Answer · answered Apr 17 '14 at 15:37

From the identity $\sin^2\theta+\cos^2\theta=1$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$, rewrite the LHS as $$ \begin{align} \frac{1}{2}(1-\cos x)&=\frac{1}{2}\left(\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-\left(\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)\right)\right)\\ &=\frac{1}{2}\left(\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-\cos^2\left(\frac x2\right)+\sin^2\left(\frac x2\right)\right)\\ &=\frac{1}{2}\left(2\sin^2\left(\frac x2\right)\right)\\ &=\sin^2\left(\frac x2\right) \end{align} $$

$$\color{blue}{\Large\text{# }\mathbb{Q.E.D.}\text{ #}}$$

Trigonometric substitution

2 Answers2