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

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.

• You're right that starting where you do then you can only prove the result when $\cos x\ne 0$. I don't know whether there is another identity you could start with, but I think not. Commented Mar 12 at 15:39
• Your identity holds for all real $\theta$, so the expression in parentheses vanishes except possibly where $\cos\theta = 0$. How best to handle the "missing points" depends what tools you have. For example, do you know that a derivative cannot have a removable discontinuity, or do you have the mean value theorem? Commented Mar 12 at 15:56
• @AndrewD.Hwang Hi, I'm aware that a function differentiable at $c$ is continuous at $c$ and the mean value theorem. Please let me know how I'm supposed to use these facts to handle this. Thanks! Commented Mar 12 at 16:08
• For example, math.stackexchange.com/questions/1833147/… Commented Mar 12 at 16:20
• @AndrewD.Hwang Thank you very much! I have read your answer in the post (I assume you meant me to read that only so I didn't go through the other answers carefully yet). It seems you were suggesting that $(\cos\theta)'$ must be $-\sin\theta$ for $\cos\theta \neq 0$, and since $\cos\theta$ is continuous on $\mathbb{R}$, it follows that its derivative is also continuous on $\mathbb{R}$, so there must not be jumps or holes in the derivative function when $\cos\theta = 0$, and thus $(\cos\theta)'$ must be $-\sin\theta$ for all real $\theta$. Please let me know if that's what you meant! Thanks! Commented Mar 12 at 17:59

You can use that $$\sin \theta + \cos \theta=\sqrt 2\sin (\theta + \frac{\pi}{4})$$ Therefore $$\cos \theta + (\cos \theta)'=\sqrt 2\cos (\theta + \frac{\pi}{4})=\sqrt 2 \left(\cos \theta \sin \frac{\pi}{4} - \sin \theta \cos \frac{\pi}{4}\right) =\cos \theta - \sin \theta \implies \\ (\cos \theta)' = -\sin \theta$$
I was inspired by Vasili's answer, and came up with an even simpler approach. Use $$\cos(x)=\sin(x+\pi/2)$$.
Then: $$\cos'(x) = \sin'(x+\pi/2) = \cos(x+\pi/2) = \sin(x+\pi) = -\sin(x)$$