Hi guysl! I've been kinda struggling with finding inverses of trig functions on different intervals.
E.g. function [MATH]y=sin^2(x/2)[/MATH] on closed interval [MATH][2 \pi, 3 \pi][/MATH]. I start with taking square root on both sides.
[MATH]\sqrt{y}=\mid sin(x/2) \mid [/MATH] and it's [MATH]\sqrt{y}=- sin(x/2) [/MATH] because sin(x/2) is negative on that interval.
But then to get rid of sin I have to apply arcsin on both sides. But arcsin is defined on [-pi/2,pi/2]. I have no idea how I could shift this function and get inverse.
Please, any tips or ideas how to generally solve such exercises? I'd be thankful from the bottom of my heart.
Have a great day.
E.g. function [MATH]y=sin^2(x/2)[/MATH] on closed interval [MATH][2 \pi, 3 \pi][/MATH]. I start with taking square root on both sides.
[MATH]\sqrt{y}=\mid sin(x/2) \mid [/MATH] and it's [MATH]\sqrt{y}=- sin(x/2) [/MATH] because sin(x/2) is negative on that interval.
But then to get rid of sin I have to apply arcsin on both sides. But arcsin is defined on [-pi/2,pi/2]. I have no idea how I could shift this function and get inverse.
Please, any tips or ideas how to generally solve such exercises? I'd be thankful from the bottom of my heart.
Have a great day.