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How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?

Answer 1

Find #tan ((5pi)/12)# and sin ((5pi)/12)
Answer: #+- (2 +- sqrt3) and +- sqrt(2 + sqrt3)/2#

Call tan ((5pi/12) = t. Use trig identity: #tan 2a = (2tan a)/(1 - tan^2 a)# #tan ((10pi)/12) = tan ((5pi)/6) = - 1/(sqrt3) = (2t)/(1 - t^2)# #t^2 - 2sqrt3t - 1 = 0#

#D = d^2 = b^2 - 4ac = 12 + 4 = 16 #--> #d = +- 4#

#t = tan ((5pi)/12) = (2sqrt3)/2 +- 4/2 = 2 +- sqrt3#

Call #sin((5pi)/12) = sin y# Use trig identity: #cos 2a = 1 - 2sin^2 a# #cos ((10pi)/12) = cos ((5pi)/6) = (-sqrt3)/2 = 1 - 2sin^2 y# #sin^2 y = (2 + sqrt3)/4# #sin y = sin ((5pi)/12) = +- sqrt(2 + sqrt3)/2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the exact values of tan(5pi/12) &amp; sin(5pi/12) using the half angle formula?

How do you find the exact values of tan(5pi/12) & sin(5pi/12) using the half angle formula?