How do you find the value of #tan15# using the half angle?

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Answer 1 · 2016-06-24T15:03:56

#tan(theta/2) = +-sqrt((1 - costheta)/(1 + costheta))#

Since we are looking for #tan15^@#, #theta = 30^@#.

#tan15^@ = +-sqrt((1 - sqrt(3)/2)/(1 + sqrt(3)/2))#

#tan15^@ = +-sqrt(((2 - sqrt(3))/2)/((2 + sqrt(3))/2))#

#tan15^@ = +-sqrt((2 - sqrt(3))/(2 + sqrt(3)#

By calculator, this can be simplified to #2 - sqrt(3)#

Since tangent is positive in quadrant #1#, #tan15^@ =2 - sqrt(3)#.

Hopefully this helps!