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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.

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    $\begingroup$ Why are you setting $f'(x) = 1$????? $\endgroup$ – fleablood Nov 7 '18 at 17:36
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    $\begingroup$ Note that $D(1) = 0$. So the right hand side of your equations for $f'$ are messed up. $\endgroup$ – copper.hat Nov 7 '18 at 17:37
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    $\begingroup$ You aren't supposed to set anything to anything. Where did you get that idea from? $\endgroup$ – fleablood Nov 7 '18 at 17:45
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    $\begingroup$ 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. $\endgroup$ – quid Nov 8 '18 at 21:09
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    $\begingroup$ Meta discussion: math.meta.stackexchange.com/a/29354/10513 $\endgroup$ – user1729 Feb 5 at 11:28
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You have that $f'(x)=0$, so $f(x)=c$. Now substitute a "nice" value into $f$ to get the value of $c$.

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  • $\begingroup$ 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? $\endgroup$ – Aning1947 Nov 7 '18 at 17:37
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$f(x)$ 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: $$ f(x)=1 \ \forall x\in \mathbb R$$

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