# Thread: show that arctanx = arcsin(x/sqrt (1+x^2)

1. ## show that arctanx = arcsin(x/sqrt (1+x^2)

also prove that

arccosx = 2arccos sqrt ( (x+1)/2 )

for the first one i let the LHS be y but didnt know where to go next

for the second one i cosined both sides but am still unsure

2. $\displaystyle y = \arctan (x)\; \Rightarrow \;\tan (y) = x\; \Rightarrow \;\sin (y) = ?$

3. Hello, b0mb3rz!

Prove that: .$\displaystyle \arccos x \:= \:2\arccos\left(\sqrt{\tfrac{x+1}{2}}\right)$
Let $\displaystyle \theta \:=\:\arccos\left(\sqrt{\tfrac{x+1}{2}}\right)\qua d \Rightarrow\quad \cos\theta \:=\:\sqrt{\tfrac{x+1}{2}}$ .[1]

The right side is: .$\displaystyle y \:=\:2\theta$

Take the cosine of both sides:

. . $\displaystyle \cos y \:=\:\cos(2\theta) \quad\Rightarrow\quad \cos y \:=\:2\cos^2\!\theta - 1$ .[2]

Substitute [1] into [2]: .$\displaystyle \cos y \;=\;2\left(\sqrt{\tfrac{x+1}{2}}\right)^2-1 \;=\;x$

We have: .$\displaystyle x \:=\:\cos y$

. . Then: .$\displaystyle \arccos x \:=\:\arccos(y) \:=\:\arccos(2\theta)$

Therefore: .$\displaystyle \arccos x \;=\;2\arccos\left(\sqrt{\tfrac{x+1}{2}}\right)$