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.
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.
2 answers
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
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