How do you evaluate the expression #cos(u+v)# given #cosu=4/7# with #0<u<pi/2# and #sinv=-9/10# with #pi<v<(3pi)/2#?

Question

1853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-31T19:51:19

#(9sqrt(33)-4sqrt(19))/70#

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

let's calculate (1)sin u and (2)cos v:

(1) #sinu=+-sqrt(1-cos^2u)#;

in the first quadrant #(0< u < pi/2) sinu>0#, then

#sinu=+sqrt(1-(4/7)^2)=sqrt(1-16/49)=sqrt((49-16)/49)=sqrt(33)/7#

(2) #cos v=+-sqrt(1-sin^2v)#

in the third quadrant #(pi < v < (3pi)/2) cosv<0#, then

#cosv=-sqrt(1-(-9/10)^2)=-sqrt(1-81/100)=-sqrt((100-81)/100)=-sqrt(19)/10#

Then

#cos(u+v)=cosucosv-sinusinv=#
#=4/7(-sqrt(19)/10)-sqrt(33)/7(-9/10)=#
#=-(4sqrt(19))/70+(9sqrt(33))/70#
#=(9sqrt(33)-4sqrt(19))/70#