Find the derivative of $\arctan \left( \cos x\over1+\sin x \right)$.

Question

Find the derivative with respect to $x$ of $$\arctan \left( \cos x\over1+\sin x \right).$$

I tried solving this problem by using trigonometric functions of submultiple of numbers (formulas of $\sin x$ and $\cos x$) but they didn't help.

Answer 1

Hint:

$$\frac{\cos x}{1+\sin x} = \frac{\sin(\frac\pi2 -\frac x2 )}{1+\cos(\frac\pi2-\frac x2) } = \frac{2\sin(\frac\pi4-\frac x2)\cos(\frac\pi4-\frac x2)}{2\cos^2(\frac\pi4-\frac x2)} = \tan(\frac\pi4-\frac x2).$$

Answer 2

It's $$\frac{1}{1+\left(\frac{\cos{x}}{1+\sin{x}}\right)^2}\cdot\left(\frac{\cos{x}}{1+\sin{x}}\right)'.$$ Can you end it now?

Also, you can use $$\frac{\cos{x}}{1+\sin{x}}=\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right).$$

Answer 3

Using https://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml

$y=\arctan\dfrac{\cos x}{1+\sin x}=\arctan\dfrac{1-t^2}{(1+t)^2}$ where $t=\tan\dfrac x2$

If $1+t\ne0$

$y=n\pi+\dfrac\pi4-\dfrac x2$ where $n$ is an integer such that $-\dfrac\pi2<y<\dfrac\pi2$