sinx=14, x in quadrant II. Find the value of sinx2
sinx=14,x in 2nd quadrant
⇒π2<x<π and cos is negative in 2nd quadrant
Using 1−cosx=2sin2x2⇒sinx2=±√1−cosx2
We get, sinx2=±
⎷1−(−√154)2=±√4+√158
As π2<x<π⇒π4<x2<π2 and sin is positive in 1st quadrant
∴sinx2=√8+2√154
Using 1+cosx=2cos2x2⇒cosx2=±√1+cosx2
We get cosx2=±
⎷1+(−√154)2=±√4−√158=±√8−2√154
As π2<x<π⇒π4<x2<π2 and cos is positive in first quadrant
∴cosx2=√8−2√154
Using cosx=1−tan2x21+tan2x2⇒tanx2=±√1−cosx1+cosx
We get tanx2=±
⎷1−(−√154)1+(−√154)=±√4+√154−√15
As π2<x<π⇒π4<x2<π2 and tan is positive in first quadrant
∴tanx2=4+√15