Answer to Question #139918 in Trigonometry for Michael

Question

If sinx = 4/5 and siny= -12/13 , 0<x<Pi over 2, 3pi over 2 < y < 2pi, evaluate

a) cos (x+y)
b) tan (x-y)

Answer 1

$\cos a = \pm \sqrt{1 - sin^2 a}, \tan a = \sin a / \cos a$

Precalculate:

$\cos x = \sqrt{1 - (4/5)^2} = 3/5, \tan x = 4/3$

$\cos y = \sqrt{1 - (-12/13)^2} = 5/13, \tan y = -12/5$

Answer:

$\begin{aligned}a) \cos(x+y) &= \cos x \cdot \cos y - \sin x \cdot \sin y \\ &= 3/5 \cdot 5/13 - 4/5 \cdot (-12/13) \\ &= 63/65\end{aligned}$

$\begin{aligned} b) \tan(x-y) &= (\tan x - \tan y) / (1 + \tan x \cdot \tan y) \\ &= (4/3 - (-12/5)) / (1 + 4/3 \cdot (-12/5)) \\ &= 56 / (15 - 48 ) = -56 / 33\end{aligned}$