# precal

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

• precal -

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

• precal -

Thank you!

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