How do you find the exact values of #costheta# and #tantheta# when #sintheta=1/2#?

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Answer 1 · 2017-01-07T10:46:32

#costheta=+-sqrt(3)/2#
#tantheta=+-sqrt(3)/3#

Since

#costheta=+-sqrt(1-sin^2theta)#

you get

#costheta=+-sqrt(1-1/4)=+-sqrt(3)/2#

Since

#tantheta=sintheta/costheta#,

you get

#tantheta=(1/cancel2)/(+-sqrt(3)/cancel2)#

The #tantheta=+-1/sqrt(3)=+-sqrt(3)/3#