Question

How do you evaluate `sin^-1(cos((3pi)/4))` without a calculator?

Answer 1

`sin^-1(cos((3pi)/4))=-pi/4`

Explanation:

Let ,

`X=sin^-1(cos((3pi)/4))`

`X=sin^-1(cos(pi-pi/4))`

`X=sin^-1(-cos(pi/4))....to[because2^(nd)Quadrant]`

`X=sin^-1(-1/sqrt2)....to[becausesin^-1(-x)=-sin^-1x]`

`X=-sin^-1(1/sqrt2)`

`X=-pi/4`

Answer 2

`- pi/4`

Explanation:

`cos theta = sin ( pi/2 - theta ), theta in ( -oo, oo ) radian`

`sin^(-1) cos ( 3/4pi ) = sin^(-1)sin (pi/2 -3/4pi)`

`= sin^(-1)sin(-pi/4)`

`= -pi/4`.

For related information:

Had it been

`sin^(-1) cos (- 5/4pi)`

`= sin^(-1)sin ( pi/2 + 5/4pi)`

`= sin^(-1)sin (9pi/4)`

`ne 9pi/4`, as it is `notin [ - pi/2, pi/2]`.

The calculators' answer is `- 45^o`

There is need for the piecewise bijective definition

`(sin)^(-1)x = kpi + (-1)^k sin^(-1)x, k = 0, +- 1, +- 2, +- 3, ...`.

Here, in this example,

`(sin)^(-1) sin ( 9/4pi) = 9/4pi`,

in accordance with `f^(-1)f(x) = x`