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#?

1 Answer
Sep 27, 2017

see below

Explanation:

Use the composite argument formula #cos (u-v)=cosucosv-sinusinv#.

But first we need to find #cos u and sin v#. Since #u# is in Quadrant II and #sin u= 3/5=(opposite)/(hypote n use# we can find the adjacent side by using pythagorean theorem.

That is,
#a^2=c^2-b^2#

#a=sqrt(5^2-3^2)=sqrt(25-9)=sqrt16=4# but since #u# is in quadrant two then #a=-4#. Hence, #cos u=-4/5#.

Likewise, we need to find #sin v# and #v# is in Quadrant III. So, to find the missing side from # cos v= -5/6=(adjacent)/(hypote n use) #
we use pythagorean theorem to find the length of the opposite side

#o=sqrt(6^2-(-5)^2)=sqrt(36-25)=sqrt11. # Since, #v# is in Quadrant III then #o=-sqrt11# and #sin v=-sqrt11/6#

Therefore,

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

#=(-4/5)(-5/6)-(3/5)(-sqrt11/6)#

#=((-2cancel4) /cancel5)(-cancel5/(3cancel6)-(cancel3/5)(-sqrt11/(2cancel6))#

#=2/3 + sqrt11/10#

#:.=(20+3sqrt11)/30#