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

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Answer 1 · 2017-01-18T08:24:05

The answer is #S={-56/65,-16/65}#

We use
#cos^2x+sin^2x=1#

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

#sinu=5/13#

Therefore,

#cos^2u=1-sin^2u=1-(5/13)^2=1-25/169=144/169#

#cosu=sqrt(144/169)=12/13#

#cosv=-3/5#

Therefore,

#sin^2v=1-cos^2v=1-(-3/5)^2=1-9/25=16/25#

#sinv=sqrt(16/25)=+-4/5#

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

or

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