How do you evaluate arcsin(sin((7pi)/3))?

1 Answer
Apr 13, 2016

arcsin(sin((7pi)/3))=(pi/3)

Explanation:

If arcsinx=theta

then sintheta=x.

Let theta=((7pi)/3) and sin((7pi)/3)=x

Then arcsin(sin((7pi)/3))=arcsinx=theta=(7pi)/3

However, in case of trigonometric ratios, for each x, their could be multiple vales of theta (such as 2npi+theta) e.g. not only sin((7pi)/3)=x, but so is for sin(pi/3), sin(-5pi/3), sin(13pi/3) etc.

Hence, the range for theta=arcsinx is limited to [-pi/2,pi/2]

Hence arcsin(sin((7pi)/3))=(pi/3)