How do you find the value of #sin ((11pi)/12)#?

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How do you find the value of #sin ((11pi)/12)#?

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Answer 1 · 2016-09-18T18:39:24

#sqrt3/2#

Explanation:

Trig table and unit circle -->
#sin ((11pi)/12) = sin (-pi/12 + (12pi)/12 = sin (-pi/12 + pi) =#
#= sin (pi/12)#.
Use trig identity;
#2sin^2 = 1 + cos pi/6 = 1 + 1/2 = 3/2#
#sin^2 (pi/12) = 3/4#
#sin (pi/12) = sqrt3/2# (because sin (pi/12) is positive)
Finally, #sin ((11pi)/12) = sin (p/12) = sqrt3/2#

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How do you find the value of #sin ((11pi)/12)#?

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