# How do you simplify the expression cos(arctan 2x - arcsin x)?

Aug 25, 2016

$\frac{\sqrt{1 - {x}^{2}} + 2 {x}^{2}}{\sqrt{1 + 4 {x}^{2}}}$

#### Explanation:

Let $a = a r c \tan 2 x \in Q 1 \mathmr{and} Q 4$, wherein cosine is positive.

Then, $\tan a = 2 x , \sin a = \frac{2 x}{\sqrt{1 + 4 {x}^{2}}} \mathmr{and} \cos a = \frac{1}{\sqrt{1 + 4 {x}^{2}}}$

Let $b = a r c \sin x \in Q 1 \mathmr{and} Q 4$, wherein cosine is positive.

Then, $\sin b = x \mathmr{and} \cos b = \sqrt{1 - {x}^{2}}$.

Now, the given expression is

$\cos \left(a - b\right)$

$= \cos a \cos b + \sin a \sin b$

$= \left(\frac{1}{\sqrt{1 + 4 {x}^{2}}}\right) \left(\sqrt{1 - {x}^{2}}\right) + \left(\frac{2 x}{\sqrt{1 + 4 {x}^{2}}}\right) \left(x\right)$

$= \frac{\sqrt{1 - {x}^{2}} + 2 {x}^{2}}{\sqrt{1 + 4 {x}^{2}}}$.

See how it works.

Let $x = \frac{1}{2}$. Then $a = a r c \sin \left(\frac{1}{2}\right) = \frac{\pi}{6} \mathmr{and} b = a r c \tan 1 = \frac{\pi}{4}$

$\cos \left(a - b\right) = \cos \left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

When $x = \frac{1}{2}$,

$\frac{2 {x}^{2} + \sqrt{1 - {x}^{2}}}{\sqrt{1 + 4 {x}^{2}}}$

#=(1/2+sqrt3/2)/sqrt(1+1)@

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$