(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
f) $\left|\cos\dfrac{x}{2}\right|=\sqrt{\dfrac{1+\cos x}{2}}$
g) $\left|\sin\dfrac{x}{2}\right|=\sqrt{\dfrac{1-\cos x}{2}}$
This is for (f): Since this is a half-angle identity I replace $x$ with $\frac{\pi}{2}$. And I'll use (C). $\cos(\frac{\pi}{2}+\frac{\pi}{2})=\cos\frac{\pi}{2}\cos\frac{\pi}{2}-\sin\frac{\pi}{2}\sin\frac{\pi}{2}\Rightarrow \cos2\frac{\pi}{2}=\cos^2\frac{\pi}{2}-\sin^2\frac{\pi}{2}$
Using power reduction identity of: $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$ yields $\cos2\frac{\pi}{2}=\dfrac{1+\cos2\frac{\pi}{2}}{2}$.
I do not believe this is correct because $\cos^2\theta\ne \cos2\theta$. Please help, but no answers.
Best Answer
I know from an earlier question of yours that you are familiar with the identities $$\cos 2w =2\cos^2 w-1=1-2\sin^2 w,\tag{$1$}$$ which can be derived fairly quickly from (C). (Yes, I have changed the name of the variable. That is deliberate.)
Now let $w=\frac{x}{2}$. Then the identities $(1)$ can be rewritten as $$\cos x=2\cos^2 \frac{x}{2}-1=1-2\sin^2 \frac{x}{2}.$$ (We are replacing $w$ by $\frac{x}{2}$. So $2w=x$.)
Look first at the identity $\cos x=2\cos^2 \frac{x}{2}-1$. This can be rewritten as $1+\cos x=2\cos^2 \frac{x}{2}$, and then as $\cos^2\frac{x}{2}=\frac{1+\cos x}{2}$.
Take the square root of both sides. We get $$\sqrt{\frac{1+\cos x}{2}}=\left|\cos \frac{x}{2}\right|.$$ Here we used the general fact that $\sqrt{a^2}=|a|$.
The other identity is proved the same way. From $\cos x=1-2\sin^2\frac{x}{2}$ we get $2\sin^2\frac{x}{2}=1-\cos x$. Divide both sides by $2$ and take the square root.