Question

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

Answer 1

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

Explanation:

Suppose that, `arc tan 2x=alpha, and, arc cos x=beta`

`rArr tan alpha=2x, and, cos beta=x`.

For the sake of brevity, we assume that `0<=x<=1`, so that,

`0<= alpha, beta <=pi/2 rArr "all Trigo. Ratios are" >0`.

Now, reqd. value `=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.!