Find tan(5π12) and sin ((5pi)/12)
Answer: ±(2±√3)and±√2+√32
Call tan ((5pi/12) = t.
Use trig identity: tan2a=2tana1−tan2a
tan(10π12)=tan(5π6)=−1√3=2t1−t2
t2−2√3t−1=0
D=d2=b2−4ac=12+4=16--> d=±4
t=tan(5π12)=2√32±42=2±√3
Call sin(5π12)=siny
Use trig identity: cos2a=1−2sin2a
cos(10π12)=cos(5π6)=−√32=1−2sin2y
sin2y=2+√34
siny=sin(5π12)=±√2+√32
