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

Question

4798 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T09:24:00

#sin(u-v)=-(15+4sqrt11)/30#
We use, #color(red)(sin(A-B)=sinAcosB-cosAsinB#

We have to evaluate #sin(u-v)= ?#

Given that, #color(blue)(sinu=3/5 ,)with, pi/2 < u < color(red)( pi),where,pto#pi

#color(blue)(cosv=-5/6,)with, pi < v < (3pi)/2#

Now,

#cosu=-sqrt(1-sin^2u),# where #pi/2< u < pi toII^(nd) Quadrant#

#cosu=-sqrt(1-9/25)=-sqrt(16/25)#

#=>color(blue)(cosu=-4/5#

Also,

#sinv=-sqrt(1-cos^2v)#,where,#pi< v < (3pi)/2to III^(rd)Quadrant#

#sinv=-sqrt(1-25/36)=-sqrt(11/36)#

#color(blue)(sinv=-sqrt11/6#

So,

#sin(u-v)=sinucosv-cosusinv#

#=>sin(u-v)=(3/5)(-5/6)-(-4/5)(-sqrt11/6)#

#=>sin(u-v)=-15/30-(4sqrt11)/30#

#=>sin(u-v)=-(15+4sqrt11)/30#