cosx=−13, x in quadrant III. Find the value of sinx2,cosx2,tanx2
cosx=−13,π<x<3π2
i.e. x lies in 3rd quadrant
Using 1−cosx=2sin2x2⇒sinx2=±√1−cosx2
We get, sinx2=±
⎷1−(−13)2=±√46
As π<x<3π2⇒π2<x2<3π4 and sin is positive in 2nd quadrant
∴sinx2=√2√5
Using 1+cosx=2cos2x2⇒cosx2=±√1+cosx2
we get, cosx2=±
⎷1+(−13)2=±√13
As π<x<3π2⇒π2<x2<3π4 and cos is negative in 2nd quadrant
∴cosx2=−1√3
Using cosx=1−tan2x21+tan2x2⇒tanx2=±√1−cosx1+cosx
We get tanx2=±
⎷1−(−13)1+(−13)=±√2
As π<x<3π2⇒π2<x2<3π4 and tan is negative in 2nd quadrant
∴tanx2=−√2