How do you find the remaining trigonometric functions of $θ$ $theta$ given $sin θ = - \frac{20}{29}$ $sintheta=-20/29$ and $θ$ $theta$ terminates in QIII?

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $sin θ = - \frac{20}{29}$ $sintheta=-20/29$ and $θ$ $theta$ terminates in QIII?

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Answer 1 · 2017-04-25T13:29:59

${cos}^{2} t = 1 - {sin}^{2} t = 1 - \frac{400}{841} = \frac{441}{841}$ $cos^2 t = 1 - sin^2 t = 1 - 400/841 = 441/841$
$cos t = \pm \frac{21}{29}$ $cos t = +- 21/29$
Since t is in Quadrant 3, cos t is negative.
$cos t = - \frac{21}{19}$ $cos t = - 21/19$
$tan t = \frac{sin}{cos} = (- \frac{20}{29}) (- \frac{29}{21}) = \frac{20}{21}$ $tan t = sin/(cos) = (-20/29)(-29/21) = 20/21$
$cot t = \frac{1}{tan} = \frac{21}{20}$ $cot t = 1/(tan) = 21/20$
$sec t = \frac{1}{cos} = - \frac{19}{21}$ $sec t = 1/(cos) = - 19/21$
$csc t = \frac{1}{sin} = - \frac{29}{20}$ $csc t = 1/(sin) = - 29/20$

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $sin θ = - \frac{20}{29}$ $sintheta=-20/29$ and $θ$ $theta$ terminates in QIII?

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How do you find the remaining trigonometric functions of thetaθtheta given sintheta=-20/29sinθ=−2029sintheta=-20/29 and thetaθtheta terminates in QIII?

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $sin θ = - \frac{20}{29}$ $sintheta=-20/29$ and $θ$ $theta$ terminates in QIII?