Question

sin(sin^{-1} \frac{ -sqrt{3} }{ 2 }+cos^{-1} \frac{ 1 }{ 2 })

Answer 1

@klbala2006 Do you know the range of \(\sin^{-1} x\) ?

Answer 2

looking at it its all 4th quadrant

Answer 3

i need to evaluate this function .
@ash2326 no i dont

Answer 4

Do you know the range and domain of sin (x) ?

Answer 5

@ash2326 domain of sin(x) is (R), while its range is [-1,1]

Answer 6

Good, to have the inverse of sin function, we limit it's range
domain= range of sin x= [-1, 1]
Range = \(\large [\frac{-\pi}{2}, \frac{\pi}{2}]\)
DO you get this?

Answer 7

yes i do

Answer 8

if x is positive, then y lies in \([0, \frac {\pi}{2}]\)
if x is negative, then y lies in \([\frac{-\pi}{2}, 0)\)
Now what's \(\sin^{-1}(\frac{-\sqrt 3}{2})\) ?

Answer 9

\[-45\]

Answer 10

\[-\frac{ \pi }{ 4 }\] in radian

Answer 11

@ash2326 complete please

Answer 12

Sorry, I was away. But sin (-45) is not \(\frac{-\sqrt 3}{2}\) it's, \(\frac{-1}{\sqrt 2}\).