# How to use derivatives to prove that $f(x)=2\cos^2\left(\frac{\pi }{4}-\frac{x}{2}\right)-\sin \left(x\right)=1$?

I'm supposed to use the derivative to prove that $$f(x)=2\cos ^2\left(\frac{\pi }{4}-\frac{x}{2}\right)-\sin \left(x\right)=1$$

What I have so far is:

$$f'\left(x\right)=D\left(2\left(\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)^2\right)-D\left(\sin\left(x\right)\right)=1$$

$$f\:'\left(x\right)=4\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)\left(-\frac{1}{2}\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)-\cos\left(x\right)=1$$

$$f\:'\left(x\right)=-2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cos\left(x\right)=1$$

And by continuing this I can get all the way to zero, which would mean that $$0=1$$ and from there by taking the derivative of $$1$$ we'd get $$0=0$$, which is obviously true. But what I'm confused about is that even if the constant was any number, let's say two, the equation would still come out as true. And obviously, the original function doesn't equal two.

I'm assuming that I'm somehow supposed to get $$1$$ to the left side as well? Honestly, I've gone through the formulas so many times, and yet I can't figure out how to do it.

• Why are you setting $f'(x) = 1$????? Nov 7 '18 at 17:36
• Note that $D(1) = 0$. So the right hand side of your equations for $f'$ are messed up. Nov 7 '18 at 17:37
• You aren't supposed to set anything to anything. Where did you get that idea from? Nov 7 '18 at 17:45
• Please do not delete posts once they got an answer. If your problem is solved consider accepting an answer. Click the checkmark close to the answer post.
– quid
Nov 8 '18 at 21:09
• Meta discussion: math.meta.stackexchange.com/a/29354/10513 Feb 5 '19 at 11:28

You have that $$f'(x)=0$$, so $$f(x)=c$$. Now substitute a "nice" value into $$f$$ to get the value of $$c$$.

• So, is it enough proof to say that cause f'(x)=0, function f(x) is constant, and because the constant is 1, f(x)=1 is also true? Nov 7 '18 at 17:37

$$f(x)$$ is $$2cos^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)-sin(x) \ \ \ \ (1)$$

The derivative of $$f(x)$$ is:

$$f'(x) = 2 * 2* cos\left(\frac{\pi}{4}-\frac{x}{2}\right) * \left(-sin\left(\frac{\pi}{4}-\frac{x}{2}\right)*(-\frac{1}{2})\right) \iff$$

$$f'(x)=2cos\left(\frac{\pi}{4}-\frac{x}{2}\right) sin\left(\frac{\pi}{4}-\frac{x}{2}\right)-cos(x)=sin\left(\frac{\pi}{2}-x\right)-cos(x)=0 \ \forall x\in \mathbb R$$

where the identity $$sin(2x)=2cos(x)sin(x)$$ was used. Now, by knowing $$f'(x) = 0 \ \forall x \in \mathbb R$$ you can say $$f(x) = c \ \forall x\in \mathbb R$$ based on a consequence of the mean value theorem. If you now set $$x=0$$ to the initial formula for $$f(x), (1)$$ you get $$f(0)=1$$ and thefore: $$f(x)=1 \ \forall x\in \mathbb R$$