This is a standard question from http://www.sgforums.com/forums/2297/topics/330037
Given that tan x = 1/p and that x is not acute, p > 0, find
(i) sin(-x)
(ii) sin(90-x)
Answer:
For tan x = 1/p, we can use the triangle method to find that the magnitude of sin x = 1 / √(1 +p²)
However, because x is not acute, and 1/p > 0, x lies in the 3rd quadrant, 180deg
So for that quadrant, sin x is negative. This gives us sin x = -1 / √(1 +p²)
Since sin (-x) = -sin x,
sin (-x) = 1 / √(1 +p²)
Note: It is now positive
sin (90 - x) = cos x
From the same triangle, find that the magnitude of cos x = p / √(1 +p²)
However, since x is in the 3rd quadrant,
cos x = -p / √(1 +p²)
