Question

If cosx=4/5 and cosy=12/13 then find cps(x+y) and sin(x-y)

Answer 1

Answer:

cos x = 4/5

cos y =(12/13)

➡

sin^2x + cos ^2x = 1

we know, sin^2x= (sin x)^2

(sinx)^2 + (4/5)^2= 1

sinx=(9/25)^1/2

= 3/5

$\{\textbf{\large{sinx=3/5}}}$

➡

so the value of siny is

(siny)^2+(cosy) ^2=1

(siny)^2+(12/13)^2=1

siny= (25/169)^1/2

$\boxed{\textbf{\large{siny=5/13}}}$

therefor

cos (x+y) = cosx X cosy - sinx X siny

=(4/5 X 12/13) - (3/5 x 5/13)

= 33/65

$\boxed{\textbf{\large{cos(x+y)=33/65}}}$

sin(x-y) = sinx X cosy - cosx X siny

=(3/5x12/13)-(4/5x 5/13)

= (16/65)

$\boxed{\textbf{\large{sin(x-y) =16/65}}}$