Question

How do I simplify sin(arccos(sqrt(2)/2)-arcsin(2x))?

Answer 1

I get `sin ( arccos(sqrt{2}/2) - arcsin(2x))` `= { 2x \pm sqrt{1 - 4x^2}}/{sqrt{2}}`

Explanation:

We have the sine of a difference, so step one will be the difference angle formula,

`sin(a-b) = sin a cos b - cos a sin b`

`sin ( arccos(sqrt{2}/2) - arcsin(2x))`

`= sin arccos(sqrt{2}/2) cos arcsin(2x) + cos arccos(sqrt{2}/2) sin arcsin(2x)`

Well the sine of arcsine and the cosine of arccosine are easy, but what about the others? Well we recognize `arccos(\sqrt{2}/2)` as `\pm 45^circ`, so

`sin arccos(\sqrt{2}/2)= \pm \sqrt{2}/2`

I'll leave the `pm` there; I try to follow the convention that arccos is all inverse cosines, versus Arccos, the principal value.

If we know the sine of an angle is `2x`, that's a side of `2x` and a hypotenuse of `1` so the other side is `\sqrt{1-4x^2}`.

`cos arcsin(2x) = \pm sqrt{1-4x^2}`

Now,

`sin ( arccos(sqrt{2}/2) - arcsin(2x))`

`=\pm \sqrt{2}/2 sqrt{1-4x^2} + (sqrt{2}/2)(2x)`

`= { 2x \pm sqrt{1 - 4x^2}}/{sqrt{2}}`