Question #19e54

1 Answer
sente
Jul 17, 2016

#2sin^2(theta) = 1, theta in [0,2pi] <=> theta in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}#

Explanation:

#2sin^2(theta) = 1#

#=> sin^2(theta) = 1/2#

#=> sin(theta) = +-sqrt(1/2) = +-sqrt(2)/2#

If we look at our unit circle, we find that #sin(theta) = sqrt(2)/2# for #theta in {pi/4, (3pi)/4}# and #sin(theta) = -sqrt(2)/2# for #theta in {(5pi)/4,(7pi)/4}#. Thus, taking the union of the sets, we find the result to be

#2sin^2(theta) = 1, theta in [0,2pi] <=> theta in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}#

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