Question

Knowing that Pi/2 < x < Pi
and that 0<y< Pi/2
that sinx = 1/5 and that cos y = 1/8, find sin (x+y), sin (x-y), cos (x+y) and cos (x-y). Explain.

Answer 1

sin = y/r
cos = x/r

sinx = 1/5
cosx = -√24/5 = -2√6/5

cosy = 1/8
siny = √63/8 = 3√7/8

I'll do sin(x+y) and you can apply the formulas for the rest.

sin(x+y) = sinx cosy + cosx siny
= (1/5)(1/8)+(-2√6/5)(3√7/8)
= 1/40 - 6√42/40
= (1-6√42)/40
makes sense, since x+y is in QIII