# Thread: Derivative of sine in degrees?

1. ## Derivative of sine in degrees?

I was reading this and I could not figure out why $\displaystyle \frac{d}{dx}sin(x)=\frac{\pi}{180}cos(x)$ in degrees.

Could anyone explain this to me?

2. ## Re: Derivative of sine in degrees?

Did you consider

$\displaystyle \displaystyle \pi^c = 180^o \implies \frac{\pi^c}{180} = 1^o$

3. ## Re: Derivative of sine in degrees?

Let $\displaystyle \sin x$ and $\displaystyle \cos x$ be the regular trigonometric functions, and let $\displaystyle \mathop{\mathrm{sind}} x$ and $\displaystyle \mathop{\mathrm{cosd}} x$ be functions that return the sine and cosine, respectively, of the angle of x degrees. That is, $\displaystyle \mathop{\mathrm{sind}} x=\sin \frac{\pi x}{180}$ and $\displaystyle \mathop{\mathrm{cosd}} x=\cos \frac{\pi x}{180}$. For example, $\displaystyle \mathop{\mathrm{sind}} 45 = \sin\frac{45\pi}{180}=\sin\frac{\pi}{4}=\sqrt{2}/2$, as expected.

Then, according to the chain rule, $\displaystyle (\mathop{\mathrm{sind}} x)'=\left(\sin \frac{\pi x}{180}\right)'=\cos\frac{\pi x}{180}\cdot\frac{\pi}{180}=\mathop{\mathrm{cosd}} x\cdot\frac{\pi}{180}$.

4. ## Re: Derivative of sine in degrees?

Oh my goodness that makes so much sense. Thank you emakarov.