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

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

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Answer 1 · 2017-03-09T20:08:50

$cos θ = \frac{\sqrt{8}}{3}$ $costheta=sqrt8/3$

Explanation:

$sin θ = \frac{opp}{hyp}$ $sintheta="opp"/"hyp"$ , in this case $\frac{1}{3}$ $1/3$ .

Given these values, we can work out the $adj$ $"adj"$ side of our imaginary triangle to work out $cos θ$ $costheta$ which equals $\frac{adj}{hyp}$ $"adj"/"hyp"$ .

$adj = \sqrt{{hyp}^{2} - {opp}^{2}} - \sqrt{3^{2} - 1^{2}} = \sqrt{8}$ $"adj"=sqrt("hyp"^2-"opp"^2)-sqrt(3^2-1^2)=sqrt8$

$cos θ = \frac{adj}{hyp} = \frac{\sqrt{8}}{3}$ $costheta="adj"/"hyp"=sqrt8/3$ .

Another way we can work this out is using the trigonometric identity:

${sin}^{2} A + {cos}^{2} A \equiv 1$ $sin^2A+cos^2A-=1$

$cos A \equiv \sqrt{1 - {sin}^{2} A}$ $cosA-=sqrt(1-sin^2A)$

$cos θ = \sqrt{1 - {(\frac{1}{3})}^{2}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}$ $costheta=sqrt(1-(1/3)^2)=sqrt(8/9)=sqrt8/3$

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

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Explanation:

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