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Sum and Difference Formula

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Find the exact value of the trigonometric function given that sin u = 12/13 and cos v = -4/5

(Both sin u and cos v are in Quadrant 2)

cos(u - v)

Nov 16, 2018
edited by Ruublrr  Nov 16, 2018

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xxxx

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Nov 16, 2018
edited by Rom  Nov 16, 2018
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cos(u-v)=cos(u)cos(v)+sin(u)sin(v)

given sin(u)=12/13 , cos(v)=-4/5

The given inforamtion is not enough to find the unique answer.

Assume cos(u)=5/13,sin(v)=3/5

cos(u-v)=-4/5*5/13+3/5*12/13=16/65

Assume cos(u)=5/13,sin(v)=-3/5

cos(u-v)=-4/5*5/13+(-3)/5*12/13=-56/65

assmue cos(u)=-5/13,sin(v)=3/5

cos(u-v)=-4/5*(-5)/13+12/13*3/5=56/65

assume cos(u)=-5/13, sin(v)=-3/5

cos(u-v)=-4/5*(-5)/13+12/13*(-3)/5=-16/65

https://faculty.atu.edu/mfinan/trigbook.pdf

Nov 16, 2018
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Thanks, fiora   !!!   CPhill  Nov 16, 2018
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Not enough info to answer this

We need to know in what quadrants u and v lie.....

Reason??

sin u  = 12/13

But cos u  might be 5/13  or - 5/13

cos v = -4/5

But sin v might be 3/5  or - 3/5   Nov 16, 2018