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

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Answer 1 · 2016-11-30T13:05:55

There are two possible solutions:

#sin theta = 1/sqrt5 " "cos theta =2/sqrt5#

#sin theta = -1/sqrt5 " "cos theta =-2/sqrt5#

Pose #x=tan theta# and #y=cos theta#

As #tan theta = 1/2#

#x/y=1/2#

and from the fundamental trigonometric identity:

#x^2+y^2=1#

Substituting #y# from the first equation:

#y=2x#

#x^2+(2x)^2=1#

#5x^2=1#

#x=+-1/sqrt5#
#y=+-2/sqrt5#