Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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.

share|cite|improve this question
If you have the power reduction identity, can't you just take the square root of both sides? I saw you had a similar question. What other parts to this did you have to do? Maybe you can use the result from a previous part in your proof. – Mike Aug 10 '12 at 8:05
But the P.R. identity states that in order to get $\dfrac{1+\cos2\theta}{2}$, it must come from $\cos^2\theta$. But the O.P. has $\cos2x$ which are not the same. – Austin Broussard Aug 10 '12 at 8:08
"The O.P."? Aren't you the O.P.? The power reduction formula is an equation to be manipulated. If you can't take square roots of both sides of an equation, how else are you going to get a radical in there? – Mike Aug 10 '12 at 8:29
up vote 1 down vote accepted

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.

share|cite|improve this answer
So, you're saying that $\cos2\frac{\pi}{2} \equiv 2\cos^2 \frac{\pi}{2}-1$? – Austin Broussard Aug 10 '12 at 8:17
That's true, but I did not explicitly say that, and it is not needed for the solution. The calculations you made involving $\pi$ are not really useful for the solution. In the post, I derived the result you want in detail from the identities for $\cos 2w$ that we discussed at length earlier. – André Nicolas Aug 10 '12 at 8:21

Note that the identity $$\cos^2\theta=\frac{1+\cos 2\theta}2$$ is equivalent to (f), so you shouldn’t be using it: your argument will necessarily be circular. Using (C) is fine, however: just set apply it to $\cos\left(\frac{x}2+\frac{x}2\right)$. You’ll also need the Pythagorean identity $\sin^2x+\cos^2x=1$.

share|cite|improve this answer
I see what you're saying! – Austin Broussard Aug 10 '12 at 8:10
Based on the question, he shouldn't be using the power reduction identity at all considering it isn't part of A through D which the question allows him to use. If however, the power reduction formula was proven in parts a through e, it should be fair game. – Mike Aug 10 '12 at 8:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.