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

Question

11863 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T11:19:29

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

Suppose that, $a r c tan 2 x = α, and, a r c cos x = β$ $arc tan 2x=alpha, and, arc cos x=beta$

$\Rightarrow tan α = 2 x, and, cos β = x$ $rArr tan alpha=2x, and, cos beta=x$ .

For the sake of brevity, we assume that $0 \leq x \leq 1$ $0<=x<=1$ , so that,

$0 \leq α, β \leq \frac{π}{2} \Rightarrow all Trigo. Ratios are > 0$ $0<= alpha, beta <=pi/2 rArr "all Trigo. Ratios are" >0$ .

Now, reqd. value $= sin (α - β)$ $=sin (alpha-beta)$

$=sin alphacos beta- cos alphasin beta...........(star)$

$tan alpha=2x rArr sec^2 alpha=tan^2 alpha+1=4x^2+1$

$rArr cos alpha=1/(+sqrt(4x^2+1)).....(1)$ .

$tan alpha=2x rArr sin alpha/cos alpha=2x$

$rArr sin alpha=2xcos alpha=(+2x)/sqrt(4x^2+1)..........(2)$

Also, $cos beta=x rArr sin beta=+sqrt(1-x^2)..........(3)$

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

$"The Reqd. Value"=(2x^2)/sqrt(4x^2+1)-sqrt(1-x^2)/sqrt(4x^2+1)$

$=(2x^2-sqrt(1-x^2))/sqrt(4x^2+1)$ .

Enjoy Maths.!

How do you simplify the expression sin(arctan 2x - arccos x)sin(arctan2x−arccosx)sin(arctan 2x - arccos x)?