Assume that $\frac{d}{d\theta}\sin\theta = \cos\theta$. Use implicit differentiation to prove that $\frac{d}{d\theta}\cos\theta = - \sin\theta$

Question

Assume that $\frac{d}{d\theta}\sin\theta = \cos\theta$. Use implicit differentiation to prove that $\frac{d}{d\theta}\cos\theta = - \sin\theta$

Here's my attempt:

$(\sin\theta)^2 + (\cos\theta)^2 = 1$

Differentiate both sides and we have:

$2\sin\theta(\frac{d}{d\theta}\sin\theta) + 2\cos\theta(\frac{d}{d\theta}\cos\theta) = 0$

Since $\frac{d}{d\theta}\sin\theta = \cos\theta$, it follows that:

$\cos\theta(\sin\theta + \frac{d}{d\theta}\cos\theta) = 0$

Now I'm stuck. We can't just conclude that $(\sin\theta + \frac{d}{d\theta}\cos\theta) = 0$ since it's possible that $\cos\theta = 0$ while the other expression isn't.

Please advise!

Answer 1

You can use that $$\sin \theta + \cos \theta=\sqrt 2\sin (\theta + \frac{\pi}{4}) $$ Therefore $$\cos \theta + (\cos \theta)'=\sqrt 2\cos (\theta + \frac{\pi}{4})=\sqrt 2 \left(\cos \theta \sin \frac{\pi}{4} - \sin \theta \cos \frac{\pi}{4}\right) =\cos \theta - \sin \theta \implies \\ (\cos \theta)' = -\sin \theta $$

Answer 2

I was inspired by Vasili's answer, and came up with an even simpler approach. Use $\cos(x)=\sin(x+\pi/2)$.

Then: $$\cos'(x) = \sin'(x+\pi/2) = \cos(x+\pi/2) = \sin(x+\pi) = -\sin(x)$$