How do you use $sin θ = \frac{1}{3}$ $sintheta=1/3$ to find $sec θ$ $sectheta$ ?

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Answer 1 · 2018-02-19T17:49:44

$sin θ = \frac{1}{3}$ $sintheta=1/3$

by the identity, we know, ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x +sin^2x=1$

therefore, ${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta= 1- cos^2theta$

${cos}^{2} θ = 1 - {(\frac{1}{3})}^{2}$ $cos^2theta= 1- (1/3)^2$

${cos}^{2} θ = 1 - (\frac{1}{9})$ $cos^2theta= 1- (1/9)$

${cos}^{2} θ = \frac{8}{9}$ $cos^2theta= 8/9$

$cos θ = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$ $costheta= +-sqrt(8/9)= +-(2sqrt2)/3$

now, we know, $secθ=1/costheta$

$secθ=+-3/(2sqrt2)$

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