Question #ce381

Question

263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-24T02:35:56

Given #sin x=3/5 # and #x# is acute.

We get #cosx=sqrt(1-sin^2x)=sqrt(1-9/25)=4/5#.

Again #siny=12/13 # and #y# is acute.

We get #cosy=sqrt(1-sin^2y)=sqrt(1-12^2/13^2)=5/13#

Now

(a)sin(x+y)

#=sinxcosy+cosxsiny#

#=3/5*5/13+4/5*12/13=63/65#

(b)cos(x-y)

#=cosxcosy+sinxsiny#

#=4/5*5/13+3/5*12/13=56/65#

So #sin(x-y)#

#=sqrt(1-cos^2(x-y))#

#=sqrt(1-56^2/65^2)=33/65#

Now cos(x+y)

#=cosxcosy-sinxsiny#

#=4/5*5/13-3/5*12/13=-16/65#

(c) #tan(x+y)#

#=sin(x+y)/cos(x+y)#

#=(63/65)/(-16/65)=-63/16#

(d) #cot(x-y)#

#=cos(x-y)/sin(x-y)#

#=(56/65)/(33/65)#

#=56/33#