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I'm trying to find ##\sin (\arccos x)##. I let ##\theta = \arccos x## and then use ##\sin ^2 \theta + \cos ^2 \theta = 1##, I get ##\sin (\arccos x) = \pm \sqrt{1 - x^2}##. I'm not sure whether to take the positive or negative root. On Wolfram Alpha is shows that the result is the positive root, but I'm not sure why...