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How do you use the half-angle identity to find the exact value of sin67.5°?

Answer 1

Find #sin 67^@5#

Ans: sqrt(2 + sqrt2)/2

Call sin (67.5) = sin t cos 2t = cos 135 = -sqrt2/2 (Trig Table of Special Arcs). Apply the trig identity: #cos 2t = 1 - 2sin^2 t.# #-sqrt2/2 = 1 - 2sin^2 t# #2sin^2 t = 1 + sqrt2 = (2+sqrt2)/2 # #sin^2 t = (2 + sqrt2)/4# #sin t = sqrt(2 + sqrt2)/2#. Only the positive answer is accepted since the arc 67.5 is in Quadrant I. #sin t = sin (67.5) = sqrt(2 + sqrt2)/2#.

Check by calculator> sin (67.5) = 0.92 #sqrt(2 + sqrt2)/2 = 1.85/2 = 0. 92#. OK

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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.