Question

Find for , x in quadrant II

Answer 1

As, x is in quadrant II,

π 2 <x<π π 4 < x 2 < π 2

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

It is given that,

sinx= 1 4

So,

cos 2 x=1− sin 2 x =1− ( 1 4 ) 2 =1− 1 16 = 15 16

Therefore,

cos 2 x= 15 16 cosx=± 15 4 =− 15 4

As x is in quadrant III, cosx is negative.

For sin x 2 ,

sin 2 x 2 = 1−cosx 2 = 1−( − 15 4 ) 2 = 4+ 15 8 × 2 2 sin x 2 = 8+2 15 4

Now, for cos x 2

cos 2 x 2 = 1+cosx 2 = 1+( − 15 4 ) 2 = 4− 15 8 × 2 2 cos x 2 =  8−2 15 4        ( ∵cos x 2  is positive )

For tan x 2 ,

tan x 2 = sin x 2 cos x 2 = 8+2 15 4     8−2 15 4     = 8+2 15 8−2 15

Simplify the above equation,

tan x 2 =4+ 15

Hence, the respective values of sin x 2 , cos x 2 and tan x 2 are 8+2 15 4 ,  8−2 15 4 and 4+ 15 .