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

1 Answer
Nov 1, 2015

#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}#