Question

Find `sin  x/2, cos  x/2 and tan  x/2` of the following

`tan x = -4/3`, x in quadrant II

Answer 1

Here, x is in quadrant II

i.e., `pi/2 < x <pi` 

⇒ `pi/4 < x/2 < pi/2`

Therefore, sin `x/2`, cos `x/2` and tan `x/2` are all positive.

It is given that tan x = `-4/3`

`sec^2 x = 1 + tan^2 x = 1 + (-4/3)^2 = 1 + 16/9 = 25/9`

∴ `cos^2 x = 9/25`

⇒ cos x = ± `3/5`

Since x is in quadrant 2, cosx is negative

∴ `cos x = (-3)/5`

Now, cos x = `2cos^2  x/2 - 1`

⇒ `(-3)/5 = 2 cos^2  x/2 - 1`

⇒ `2cos^2  x/2 = 2/5`

⇒ `cos^2  x/2 = 1/5`

⇒ `cos  x/2 = 1/sqrt5`

`[∵ cos x/2 "is positive"]`

∴ `cos  x/2 = sqrt5/5`

⇒ `sin^2  x/2 + cos^2  x/2 = 1`

⇒ `sin^2  x/2 + (1/sqrt5)^2 = 1`

⇒ `sin^2  x/2 = 1 - 1/5 = 4/5`

⇒ `sin  x/2 = 2 /sqrt5`

`[∵ sin  x/2 "is positive"`

∴ `sin  x/2 = (2sqrt5)/5`

tan `x/2 = (sin  x/2 (2)/(sqrt5))/(cos  x/2 (1/sqrt5)) =2`

The corresponding values of `sin  x/2, cos  x/2 tan  x/2 (2sqrt5)/5, sqrt5/5, and 2`