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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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Other answers:

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 \(\sin^{-1}(\frac{-\sqrt 3}{2})\) ?
\[-45\]
\[-\frac{ \pi }{ 4 }\] 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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