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posted by anonymous Sunday, May 23, 2010 at 5:56pm.

How do you solve: sin[arccos(-2/7)] ?

cos is - in quadrants 2 and 3 adjacent is -2, hypotenuse is 7, what is opposite? 7^2 - 2^2 = 49 - 4 = 45 = 9 * 5 sqrt(9*5) = 3 sqrt (5) so sin = +/- 3 sqrt(5) /7

Thank you!

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