# How do you simplify cos(arcsinx - arctan2x)?

Jul 14, 2016

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

#### Explanation:

Let arcsinx=A, &, arctan2x=B, so that, sinA=x, &, tanB=2x, A in [-pi/2,pi/2], B in (-pi/2,pi/2)

We will consider the case $A , B \in \left[0 , \frac{\pi}{2}\right]$, so that all trigonometric ratios will be $+ v e$. The other case can be dealt with similarly.

Now reqd. value $= \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B \ldots \ldots . \left(I\right)$

$\sin A = x \Rightarrow \cos A = \sqrt{1 - {x}^{2}}$

$\tan B = 2 x \Rightarrow {\sec}^{2} B = 1 + {\tan}^{2} B = 1 + 4 {x}^{2} \Rightarrow \sec B = \sqrt{1 + 4 {x}^{2}} \Rightarrow \cos B = \frac{1}{\sqrt{1 + 4 {x}^{2}}}$

$\tan B = 2 x \Rightarrow \cot B = \frac{1}{2 x} \Rightarrow {\csc}^{2} B = 1 + {\cot}^{2} B = 1 + \frac{1}{4 {x}^{2}} = \frac{4 {x}^{2} + 1}{4 {x}^{2}} \Rightarrow \csc B = \frac{\sqrt{1 + 4 {x}^{2}}}{2 x} \Rightarrow \sin B = \frac{2 x}{\sqrt{1 + 4 {x}^{2}}}$

Sub.ing all these in $\left(I\right)$, we have,

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

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

Hope, this is Helpful! Enjoy Maths. , &, yes, don't forget to consider

the case A,B in [-pi/2,0] !