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+2 votes

Find the remaining values for cos(θ) = -8/19, and θ in quadrant III?

asked Feb 20, 2013 in TRIGONOMETRY by abstain12 Apprentice

2 Answers

+3 votes

cos(θ) = -8/9.

Pythagorean Identities: sin2(θ) + cos2(θ) = 1.

Substitute cos(θ) = -8/9 in the Identities.

sin2(θ) + (-8/9)2 = 1

sin2(θ) + 64/361 = 1

Subtract 64/361 from each side.

sin2(θ) = 1 - 64/361

Rewrite the expression with common denominator.

sin2(θ) = [361 - 64]/361

sin2(θ) = 297/361

Take square root each side.

√[sin2(θ)] = √[297/361] = √[9·33]/√[361]

sin(θ) = 3√33/19.

Tan(θ) = sin(θ)/cos(θ) = [-8/19]/[3√33/19] = -8/(3√33).

Csc(θ) = 1/sin(θ) = 1/[3√33/19] = 19/(3√33).

Sec(θ) = 1/cos(θ) = 1/[-8/19] = 19/(-8) = -19/8.

Cot(θ) = 1/tan(θ) = 1/[-8/3√33] = (3√33)/(-8) = -(3√33)/8.

Therefore Sin(θ) = 3√33/19,Cos(θ) = -8/19, Tan(θ) = -8/(3√33),

Csc(θ) = 19/(3√33), Sec(θ) = -19/8, Cot(θ) = -(3√33)/8.

answered Feb 22, 2013 by britally Apprentice

sin(θ) = -3√33/19, cot(θ) = 8/3√33, csc(θ) = - 19/3√33, tan(θ) = 3√33/8.

0 votes

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From the pythageron therom

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Since theta lies in 3 rd quadrant y is negative.

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answered Jun 23, 2014 by david Expert

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