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Question

(i) If 0 ≤ x ≤ π and x lies in the IInd quadrant such that sin x=14. Find the values of cosx2, sinx2 and tanx2
(ii) If cos x=45 and x is acute, find tan 2x
(iii) If sin x=45 and 0<x<π2, find the value of sin 4x.

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Solution

(i) sin x=14

sinx=1-cos2x142=1-cos2x116-1=-cos2x1516=cos2xcosx=±154

Since x lies in the 2nd quadrant, cos x is negative.

Thus,

cosx=-154

Now, using the identity cosx=2cos2x2-1, we get

-154=2cos2x2-1-158=cos2x2-12cos2x2=4-158cosx2=±4-158

Since x lies in the 2nd quadrant and x2 lies in the 1st quadrant, cosx2 is positive.

cosx2=4-158

Again,

cosx=cos2x2-sin2x2-154=4-1582-sin2x2-154=4-158-sin2x2sin2x2=4+158sinx2=±4+158=4+158

Now,

tanx2=sinx2cosx2 =4+1584-158=4+154-15 =4+154+154-154+15 =4+1542-152=4+1516-15=4+15

(ii) cos x=45

sinx=1-cos2x =1-452 =1-1625 =25-1625 =925 =35

tan x=sin xcos x =3545 =34

Now,
tan 2x=2 tanx1-tan2x =2341-342 =2341-916 =32716 =247

Hence, the value of tan 2x is 247.

(iii) sin x=45 and 0<x<π2.

sin x=1-cos2 x452=1-cos2 x1625-1=-cos2 x925=cos2 xcos x=±35

Since x lies in the 1st quadrant, cos x is positive.

Thus,

cos x=35

Now,
sin 4x=2 sin 2x cos2x =22 sin x cos x1-2 sin2 x =22×45×351-2452 =224251-3225 =2242525-3225 =22425-725 =-336625

Hence, the value of sin 4x is -336625.

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