Q. 22

Find if sin x = 1/4, x in quadrant II.

Answer :

Given that x is in quadrant II


So,


90° < x < 180°


Dividing with 2 all sides


(90°/2) < x/2 < (180°/2)


45° < x/2 < 90°


lies in 1st quadrant


In 1st quadrant,


sin, cos & tan all are positive


all are positive


Given


sin x = (1/4)


We know that


cos2x = 1 - sin2x


= 1 - (1/4)2


= 1 - (1/16)


cos2x = 15/16


cos x = (√15)/4


Since x is in IInd quadrant


cos x is negative


cos x = - (√15)/4


Also,


cos 2x = 2 cos2x - 1


Replacing x with x/2


cos 2(x/2) = 2 cos2(x/2) - 1


cos x = 2 cos2(x/2) - 1


- (√15)/4 = 2 cos2(x/2) - 1


1- (√15)/4 = 2 cos2(x/2)


cos2(x/2) = [4 - (√15)]/8



lies in 1st quadrant


is positive




Now,


We know that


1 + tan2x = sec2x


Replacing x with x/2








lies in 1st quadrant


is positive in 1st quadrant



We know that-



Replacing x with x/2




Hence,


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