Find #sintheta#, #costheta#, and #tantheta# under the given conditions: #sin2theta=1/5, pi/2 <= 2theta < pi#?

#pi/2<=2theta<=pi -> pi/4<=theta<=pi/2#

Thus, all double-angle cosines must be negative due to the double angle being in the second quadrant, and all single-angle sines, cosines, and tangents must be positive due to the single angle being in a portion of the first quadrant.

Keeping that in mind, recall that

#sin^2theta+cos^2theta=1#

We may optimize this to

#sin^2(2theta)+cos^2(2theta)=1#

#sin^2(2theta)=(1/5)^2=1/25#

#cos^2(2theta)=25/25-1/25#

#cos(2theta)=+-sqrt(24)/5#

We want the negative solution.

#cos(2theta)=-2sqrt(6)/5#

Now, we want to relate these double angles to single angle trig functions.

The power-reduction identities are the way to go.

Recall

#cos^2theta=1/2(1+cos2theta)#

Thus,

#cos^2theta=1/2(1-2sqrt6/5)=1/2-sqrt6/5#

#costheta=+-sqrt(1/2-sqrt6/5)#

We want the positive answer.

#costheta=sqrt(1/2-sqrt6/5)#

Furthermore, recall

#sin^2theta=1/2(1-cos2theta)#

Then,

#sin^2theta=1/2(1-(-2sqrt6/5))#

#sin^2theta=1/2+sqrt6/5#

#sintheta=+-sqrt(1/2+sqrt6/5)#

We want the positive answer.

#sintheta=sqrt(1/2+sqrt6/5)#

#tantheta=sintheta/costheta,# so

#tantheta=sqrt(1/2+sqrt6/5)/sqrt(1/2-sqrt6/5)=sqrt((1/2+sqrt6/5)/(1/2-sqrt6/5))=sqrt(((5+2sqrt6)/cancel(10))/((5-2sqrt6)/cancel(10)))=sqrt((5+2sqrt6)/(5-2sqrt6)#