How do you evaluate $arcsin (sin 2)$ $arcsin(sin 2)$ ?

Question

15942 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-19T21:24:41

The range of arcsin is $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ $-pi/2 <= theta <= pi/2$ .

$2 > \frac{π}{2}$ $2 > pi/2$ so it lies outside the range.

Use $sin (π - θ) = sin θ$ $sin (pi-theta) = sin theta$

Then $sin 2 = sin (π - 2)$ $sin2 = sin (pi-2)$ .

$π - 2 ≅ 1.14$ $pi-2 ~= 1.14$ so $0 < π - 2 < \frac{π}{2}$ $0 < pi-2 < pi/2$

So $arcsin (sin (2)) = π - 2$ $arcsin(sin(2)) = pi-2$

How do you evaluate arcsin(sin 2)arcsin(sin2)arcsin(sin 2)?