Answer: 0
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Explanation:
The original equation is in the form a+b = 1, where
a = sin(theta)
b = cos(theta)
Square both sides of a+b = 1 to get
(a+b)^2 = 1^2
a^2+2ab+b^2 = 1
(a^2+b^2)+2ab = 1
From here notice that a^2+b^2 is sin^2+cos^2 = 1, which is the pythagorean trig identity. So we go from (a^2+b^2)+2ab = 1 to 1+2ab = 1 to 2ab = 0 to ab = 0
Therefore,
sin(theta)*cos(theta) = 0
