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Question

sinx=14, x in quadrant II. Find the value of sinx2

Solution
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sinx=14,x in 2nd quadrant
π2<x<π and cos is negative in 2nd quadrant
Using 1cosx=2sin2x2sinx2=±1cosx2
We get, sinx2=± 1(154)2=±4+158
As π2<x<ππ4<x2<π2 and sin is positive in 1st quadrant
sinx2=8+2154
Using 1+cosx=2cos2x2cosx2=±1+cosx2
We get cosx2=± 1+(154)2=±4158=±82154
As π2<x<ππ4<x2<π2 and cos is positive in first quadrant
cosx2=82154
Using cosx=1tan2x21+tan2x2tanx2=±1cosx1+cosx
We get tanx2=±   1(154)1+(154)=±4+15415
As π2<x<ππ4<x2<π2 and tan is positive in first quadrant
tanx2=4+15

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