inverse and compound angle qn (1 Viewer)

[Masaken]

how would you do this question? thanks in advance, not sure how to start and where to go

 

[cossine]

can you draw a triangle. Then consider what happens for each quadrant.

So in first quadrant 0-90 degrees, k = pi/2.

 

[Masaken]

can you draw a triangle. Then consider what happens for each quadrant.

So in first quadrant 0-90 degrees, k = pi/2.

not sure about second quadrant or third (it shouldn't be any value right because of the restriction of k?), but for fourth it would be k = -pi/2 for sin inverse? what happens afterward? (sorry but i'm not sure where we're meant to go from here even after that)

 

[tyrone97]

The question is wrong and doesn't make sense.
sin-1(x) + cos-1(x) = pi/2 for all -1<=x<=+1. So it only makes sense for the case k = pi/2, and x can be anything in the domain of sin-1 & cos-1.
(You can see that the LHS is constant by differentiating LHS and seeing that it equals zero.)

 

[cossine]

not sure about second quadrant or third (it shouldn't be any value right because of the restriction of k?), but for fourth it would be k = -pi/2 for sin inverse? what happens afterward? (sorry but i'm not sure where we're meant to go from here even after that)

Hi take a look at tyrone97 post