Question

How do you find the exact value of sinx/2, cosx/2, & tanx/2 given that `sinx= 5/13` is in Quadrant 2?

Answer 1

First find cos x by using trig identity: `cos^2 x + sin^2 x = 1.`
`cos^2 x = 1 - 25/169 = (169 - 25)/169 = 144/169`
--> cos x = -12/13. (Quadrant II)
Next, use the trig identity: `cos 2a = 2.(cos^2 a) - 1`.
`cos x = 2cos^2 (x/2) - 1` -->
`2cos^2 (x/2) = 1 - 12/13 = 1/13`-->
`cos^2 (x/2) = 1/26`
--> `cos x/2 = -1/5.1` (Quadrant II)
Next, use the trig identity: `cos x = cos^2 (x/2) - sin^2 (x/2)`-->
`sin^2 (x/2) = 1/26 + 12/13 = 25/26` -->
`sin (x/2) = 5/5.1` (quadrant II)
Find `tan (x/2) = sin (x/2)/cos (x/2) = (5/5.1)/(-1/5.1) = -5/1 = -5`