Question

How do you find the value of `sin^(-1)(sin (cos^(-1) (sin (pi/12))))`?

Answer 1

`sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12`.

Explanation:

We will use the following Rules :

`(R1) : sintheta=cos(pi/2-theta)`

`(R2) : cos^-1(costheta)=theta, theta in [o,pi]`

`(R3) : sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]`

Let us note that, by

`(R1), sin(pi/12)=cos(pi/2-pi/12)=cos(5pi/12)`

So, `cos^-1(sin(pi/12))=cos^-1(cos(5pi/12))`, where,

`5pi/12 in [0,pi]`, so, using `(R2)`, we get,

`cos^-1(cos(5pi/12))=5pi/12`

Again, as `5pi/12 in [-pi/2,pi/2]`, by `(R3)`, we have,

`sin^-1(sin(5pi/12))=5pi/12`

Finally, `sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12`.

Hope, this will be of Help! Enjoy Maths.!

Answer 2

`(5pi)/12=75^o`

Explanation:

Use `sin a = cos (pi/2-a) and,`

if `y=f(x) , x = f^(-1)y, f f^(-1)y=y and f^(-1)yf(x)=x`.

The given expression is

`sin^(-1)sin cos^(-1)sin(pi/12)`

`=sin^(-1)sin (cos^(-1)cos(pi/2-pi/12))`

`=sin^(-1)sin((5pi)/12)`

`=(5pi)/12`