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

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@klbala2006 Do you know the range of \(\sin^{-1} x\) ?
looking at it its all 4th quadrant |dw:1352871092096:dw|
i need to evaluate this function . @ash2326 no i dont

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Do you know the range and domain of sin (x) ?
@ash2326 domain of sin(x) is (R), while its range is [-1,1]
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?
yes i do
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 \(\sin^{-1}(\frac{-\sqrt 3}{2})\) ?
\[-\frac{ \pi }{ 4 }\] in radian
@ash2326 complete please
Sorry, I was away. But sin (-45) is not \(\frac{-\sqrt 3}{2}\) it's, \(\frac{-1}{\sqrt 2}\).
isnt it the same ?
What's sin 60 ?

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