Question

Find the exact value of cos(u+v) given that sine u=4/5 with u in quadrant II and sine v = -12/13 with v in quadrant IV.

Not sure how to even begin.

Answer 1

first, draw your triangles in standard position. Then it is easy to see that

cos(u) = -3/5
cos(v) = 5/13

Then recall your sum formula:

cos(u+v) = cosu cosv - sinu sinv

now just plug in your numbers.

Answer 2

sin u = 4 / 5

cos u = sqroot ( 1 - sin ^ 2 u )

cos u = sqroot [ 1 - ( 4 / 5 ) ^ 2 ]

cos u = sqroot ( 1 - 16 / 25 )

cos u = sqroot ( 25 / 25 - 16 / 25 )

cos u = sqroot ( 9 / 25 )

cos u = ± 3 / 5

In Quadrant II cos is negative so:

cos u = - 3 / 5


sin v = - 12 /13

cos v = sqroot ( 1 - sin ^ 2 v )

cos v = sqroot [ 1 - ( - 12 / 13 ) ^ 2 ]

cos v = sqroot ( 1 - 144 / 169 )

cos v = sqroot ( 169 / 169 - 144 / 169 )

cos v = sqroot ( 25 / 169 )

cos v = ± 5 / 13

In Quadrant IV cos is postive so:

cos v = 5 / 13


sin u = 4 / 5, cos u = - 3 / 5, sin v = - 12 /13, cos v = 5 / 13,

cos ( u + v ) = cos u * cos v - sin u * sin v

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

- 3 / 13 + 48 / 65 =

- 3 * 5 / ( 13 * 5 ) + 48 / 65 =

- 15 / 65 + 48 / 65 = 33 / 65


cos ( u + v ) = 33 / 65