How do you find the exact values of #sintheta# and #tantheta# when #costheta=2/5#?

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Answer 1 · 2017-02-11T15:40:02

#sin theta=sqrt21/5#
#tantheta=sqrt21/2#

We need

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

#tan theta=sintheta/costheta#

#costheta=2/5#

#sintheta=sqrt(1-cos^2theta)#

#=sqrt(1-4/25)=sqrt(21/25)#

#=sqrt21/5#

#tantheta=sintheta/costheta=(sqrt21/5)/(2/5)#

#=sqrt21/2#