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

**b0mb3rz** · March 25th 2009, 02:23 AM

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

**Plato** · March 25th 2009, 03:03 AM

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

**Soroban** · March 25th 2009, 08:10 AM

Hello, b0mb3rz!

Prove that: . $\arccos x \:= \:2\arccos\left(\sqrt{\tfrac{x+1}{2}}\right)$

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

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

Take the cosine of both sides:

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

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

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

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

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

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