# How do you simplify the expression sin(arctan 2x - arccos x)?

Aug 29, 2016

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

#### Explanation:

Suppose that, $a r c \tan 2 x = \alpha , \mathmr{and} , a r c \cos x = \beta$

$\Rightarrow \tan \alpha = 2 x , \mathmr{and} , \cos \beta = x$.

For the sake of brevity, we assume that $0 \le x \le 1$, so that,

$0 \le \alpha , \beta \le \frac{\pi}{2} \Rightarrow \text{all Trigo. Ratios are} > 0$.

Now, reqd. value $= \sin \left(\alpha - \beta\right)$

$= \sin \alpha \cos \beta - \cos \alpha \sin \beta \ldots \ldots \ldots . . \left(\star\right)$

$\tan \alpha = 2 x \Rightarrow {\sec}^{2} \alpha = {\tan}^{2} \alpha + 1 = 4 {x}^{2} + 1$

$\Rightarrow \cos \alpha = \frac{1}{+ \sqrt{4 {x}^{2} + 1}} \ldots . . \left(1\right)$.

$\tan \alpha = 2 x \Rightarrow \sin \frac{\alpha}{\cos} \alpha = 2 x$

$\Rightarrow \sin \alpha = 2 x \cos \alpha = \frac{+ 2 x}{\sqrt{4 {x}^{2} + 1}} \ldots \ldots \ldots . \left(2\right)$

Also, $\cos \beta = x \Rightarrow \sin \beta = + \sqrt{1 - {x}^{2}} \ldots \ldots \ldots . \left(3\right)$

Using $\left(1\right) - \left(3\right)$ in $\left(\star\right)$, bearing in mind that $\cos \beta = x$,

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

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

Enjoy Maths.!