Question

How do you do simplify `sin^-1(cos ((5pi)/6))`?

Answer 1

`sin^{-1}(cos(frac{5pi}{6}))=-frac{pi}{3}`

Explanation:

Since `cos(pi-x)=-cos(x)`,
`cos(frac{5pi}{6})=-cos(pi/6)`.

`cos(pi/6)=frac{sqrt{3}}{2}` can be seen from the geometry of a 30,60,90 degree triangle, which is half of an equilateral triangle. Therefore,

`cos(frac{5pi}{6})=-frac{sqrt{3}}{2}`.

Since `sin^{-1}(-x)=-sin^{-1}(x)`, and `sin^{-1}(frac{sqrt{3}}{2})=frac{pi}{3}` (which can be seen from the same 30,60,90 degree triangle),

`sin^{-1}(cos(frac{5pi}{6}))=sin^{-1}(-frac{sqrt{3}}{2})`
`=-sin^{-1}(frac{sqrt{3}}{2})=-frac{pi}{3}`