How do you evaluate the expression #cos(u+v)# given #sinu=3/5# with #pi/2<u<p# and #cosv=-5/6# with #pi<v<(3pi)/2#?

Question

4108 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-28T07:43:29

#29/30#

Since

#cos(u+v)=cosucosv-sinusinv#,

let's find #cos u and sin v# and then substitute all the numeric values in the expression:

#cos u=-sqrt(1-sin^2u)=-sqrt(1-9/25)=-sqrt(16/25)=-4/5#

(#cos u<0# in the second quadrant)

#sinv=-sqrt(1-cos^2v)=-sqrt(1-25/36)=-sqrt(9/36)=-1/2#

(#sinv<0# in the third quadrant)

now let's substitute:

#cos(u+v)=cosucosv-sinusinv=#
#=-4/cancel5(-cancel5/6)-3/5(-1/2)=#
#2/3+3/10=29/30#