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in question 1 well evaluate the following compositions in part a lets find sine inverse of sine of pi over 3. a function and its inverse should cancel each other so perhaps the answer here is pi over 3. to be careful to check this im going to draw pi over 3 on the unit circle so ill put that over here and heres my unit circle and the angle pi over 3 is 60 degrees so thats like right about there and what ill do is figure out well first whats sine of pi over 3 sine of pi over 3 is equal to the square root of 3 over 2. so sine takes pi over 3 to root 3 over 2. sine inverse should reverse this it should take us from root 3 over 2 back to pi over 3. that is true here so sine inverse of root 3 over 2 undoes this and gives us pi over 3 back but this is only true if the angle falls between negative pi over 2 and pi over 2. so sine of power 3 goes to root 3 over 2 and sine inverse goes backwards from root 3 over 2 to pi over 3 but thats only because pi over 3 falls within this range f
