Question

How do you find `sin^-1(sin ((5pi)/6))`?

Answer 1

`sin^-1(sin(5pi)/6)=pi/6`

Explanation:

`sin((5pi)/6)=1/2`, so we're really being asked for `sin^-1(1/2).`

Let

`y=sin^-1(1/2)`

Then, from the definition of arcsine,

`1/2=sin(y)`

( `y=sin^-1(a) hArr a=sin(y)` )

In other words, we want to know where the sine function returns `1/2.`

Sine is equal to `1/2` for `pi/6+2npi, (5pi)/6+2npi,` where `n` is any integer. But really, since the domain of arcsine is `[-pi/2, pi/2], pi/6` is the only valid answer.

Answer 2

`pi/6`

Explanation:

Hint:
We know that ,
`color(red)(sin^-1(sintheta)=theta,` where, `color(red)(theta in [-pi/2,pi/2]`
Even though,
`color(blue)(sin^-1(sin((5pi)/6))!=((5pi)/6),` because, `color(blue)((5pi)/6 !in [-pi/2,pi/2]`
So, we reduce `theta=(5pi)/6.` But how ? By using Addition and Subtraction Formulas for sine and cosine .Now enjoy the answer.

****Answer:****

Here,

`(5pi)/6=pi-pi/6`

`(sin^-1(sin((5pi)/6))=sin^-1(sin(pi-pi/6))`

`=sin^-1(sin(pi/6))....to`, Apply `color(red)([sin(pi-theta)=sintheta ]`

`=pi/6......to`, where, `pi/6in[-pi/2,pi/2]`