Question

sin^-1(sin(17pi/19))

Answer 1

\(\huge sin^{-1}(sin \:a)=a\)

Answer 2

so what will be sin^-1(sin(17pi/19)) =?

Answer 3

The answer is \[\frac{ 2\pi }{19}\] But I don't understand how or why.

Answer 4

is it because it's not within \[\frac{ -\pi }{ 2 } and \frac{ \pi }{ 2 }\] ?

Answer 5

yes, 17pi/19 is not in that range, so u need to add or subtract 2\(\pi\) to bring the angle in that range.

Answer 6

Did you mean just pi?

Answer 7

the arcsin is defined on the interval \((-\pi/2,\pi/2)\) so we need to do something first with this problem.
\(\sin^{_1}(\sin 17\pi/19)=\sin^{-1}(\sin 2\pi/19)=2\pi/19\)

Answer 8

since it's in QII we use theta minus pi right?

Answer 9

\(\sin \theta=\sin (\pi-\theta)\)

Answer 10

u bring it in 1st quadrant by using that formula ^
so sin(17pi/19)= sin(pi-17pi/19)

Answer 11

oops, meant pi minus theta. Thank you both!