# Why does $\frac{d}{dx} \cos ( x - \tan^{-1} ( 2 ) ) = \frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$?

The title just about sums up my question.

Wolfram|Alpha shows it to be $\frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$, while the (extremely simple) derivation I did by hand gives $-\sin(x - \tan^{-1}(2))$ (which wolfram agrees with).

I'm perfectly willing to accept that the two are equal, but I'd like to know why. What is the property or relationship between $\frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$ and $-\sin(x - \tan^{-1}(2))$ that allows you to convert from one form to another without changing the value of the expression?

I've done some looking, and my initial thoughts are that it's a property of the arc tangent, but I haven't been able to find a solid answer.

Thanks in advance for any responses.

• Consider a right triangle with height 2, base 1. Let $\theta$ be angle between hypotenuse and base. Find $\cos \theta$, $\sin \theta$, you'll get your answer – Max Payne Dec 12 '16 at 9:52

use the fact that: $$\tan^{-1}(2)=\alpha \iff \tan \alpha =2$$ and: $$\sin \alpha=\frac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}}=\frac{2}{\sqrt{5}}$$
$$\cos \alpha=\frac{1}{\sqrt{1+\tan^2 \alpha}}=\frac{1}{\sqrt{5}}$$
For now, let the term $x-\arctan(2) =\alpha$. We know the differentiation of $\cos \alpha$ equals $-\sin \alpha$. Hence we have $$\frac{d(\cos (x-\arctan(2)))}{dx} =-\sin(x-\arctan(2))$$
Now, use the formula $\sin(A-B) = \sin A\cos B-\cos A\sin B$. Here $A=x$ and $B=\arctan(2)$. In a triangle of angle $B =\arctan(2)$, we have $\sin B = \frac{2}{\sqrt{5}}$ and $\cos B =\frac{1}{\sqrt{5}}$. Using these results and applying them in the formula , gives us the derivative as $$\frac{2\cos x-\sin x}{\sqrt{5}}$$