SOLUTION: cos(tan^(-1)u-cos^(-1)v) we are supposed to write it as an algebraic expression and I just don't understand how to do it at all

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Question 1129444: cos(tan^(-1)u-cos^(-1)v) we are supposed to write it as an algebraic expression and I just don't understand how to do it at all
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


cos%28tan%5E-1%28u%29-cos%5E-1%28v%29%29
making
tan%5E-1%28u%29+=+A =>tan%28A%29+=u
cos%5E-1%28v%29=B =>cos%28B%29=v
then,
tan+%28A%29+=+u, ->recall trig. triangle
opposite_+side+=u, adjacent_+side+=+1, hypotenuse+=+sqrt%28u%5E2%2B1%29+
so,
tan%28+A%29+=opposite_+side%2Fadjacent_+side=u%2F1
since
tan+%28A%29+=sin%28A%29%2Fcos%28A%29
we will find sin%28A%29 and cos%28A%29
sin%28A%29=opposite_+side%2Fhypotenuse+=+u%2Fsqrt%28u%5E2%2B1%29+
cos%28A%29=+adjacent_+side%2Fhypotenuse+=+1%2Fsqrt%28u%5E2%2B1%29
and for

cos%28B%29=v
adjacent_+side+=v,opposite+_side+=sqrt%281-v%5E2%29, hypotenuse+=+1

cos%28B%29=adjacent+_side%2Fhypotenuse=v%2F1=v
sin%28B%29=opposite_+side%2Fhypotenuse=sqrt%281-v%5E2%29%2F1=sqrt%281-v%5E2%29

And using the cos of difference formula:

cos%28A-B%29=sin%28A%29+sin%28B%29+%2B+cos%28A%29+cos%28B%29

and since cos%28A-B%29=cos%28tan%5E-1%28u%29-cos%5E-1%28v%29%29, we have

cos%28tan%5E-1%28u%29-cos%5E-1%28v%29%29=sin%28A%29+sin%28B%29+%2B+cos%28A%29+cos%28B%29