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Question

Find sinx2,cosx2 and tanx2 for cosx=13,x in quadrant III.

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Solution

Step 1: Check sign for sinx2,cosx2 and tanx2
Given that x is in quadrant III.
180<x<270
1802<x2<2702
90<x2<135
x2 lies in 2nd quadrant.
sin is positive, cos and tan are negative.

Step 2:
cosx=2cos2x21
13=2cos2(x2)1
2cos2(x2)=113
2cos2(x2)=23
cos(x2)=±13
But cosx2 is negative
cosx2=13

Step 3:
Solve for the value of sinx2
sin2x+cos2x=1 Replacing x by x2
sin2x2+cos2x2=1
sin2x2+13=1sin2x2=23
sinx2=±23×33sinx2=±63
But sinx2 is positive
sinx2=63

Step 4:
tanx=sinxcosx Replacing x by x2
tanx2=sinx2cosx2
tanx2=6313=63×31=3×23×31
tanx2=2
sinx2=63,cosx2=33 and tanx2=2

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