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

Question

9823 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-01T04:16:43

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

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),

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

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