How do you find the exact value of #sin(u+v)# given that #sinu=5/13# and #cosv=-3/5#?

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Answer 1 · 2017-04-08T05:59:22

#-63/65# or #33/65#

Since

#sin(u+v)=sinucosv+cosusinv#,

you would get #cosu# and #sinv# before applying it:

#cosu=+-sqrt(1-sin^2u)=+-sqrt(1-25/169)=+-sqrt(144/169)=+-12/13#

and

#sinv=+-sqrt(1-cos^2v)=+-sqrt(1-(-3/5)^2)=+-sqrt(16/25)=+-4/5#

Then

#sin(u+v)=sinucosv+cosusinv=5/13*(-3/5)+-12/13*(+-4/5)=-15/65+-48/65#
Then

#sin(u+v)=(-15-48)/65=-63/65#

or

#sin(u+v)=(-15+48)/65=33/65#