Question 169985: 5.1.6. what are all of the trigonomic values given that cot=-3 and sin= square root 10/10
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! what are all of the trigonomic values given that cot=-3 and sin= square root 10/10
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Keep in mind that x^2 + y^2 = r^2
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cot=-3 = -3/1
Since cot = x/y, x = -3 and y = 1
Therefore r = sqrt(-3)^2 + 1^2) = sqrt(10)
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Then:
sin = y/r = 1/sqrt(10) = sqrt(10)/10
cos = x/r = -3/sqrt(10)
tan = y/x = -1/3
csc = r/y = sqrt(10)
sec = r/x = -sqrt(10)/3
cot = -3
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Cheers,
Stan H.
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