Masaken
Unknown Member
how would you do this question? thanks in advance, not sure how to start and where to go
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inverse and compound angle qn (1 Viewer)
- Thread starter Masaken
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how would you do this question? thanks in advance, not sure how to start and where to go
Masaken
Unknown Member
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)
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.)
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.)
Hi take a look at tyrone97 post