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mars1129 [50]
3 years ago
10

If sin θ = 1/3 and tan θ < 0, what is the value of cos θ?

Mathematics
1 answer:
Mashcka [7]3 years ago
8 0
\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad&#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\quad &#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}\\\\&#10;-------------------------------\\\\&#10;sin(\theta )=\cfrac{1}{3}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby &#10;\begin{array}{llll}&#10;\textit{so, let's use the pythagorean}\\&#10;\textit{theorem to get the adjacent}&#10;\end{array}

\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{3^2-1^2}=a\implies \pm\sqrt{8}=a\implies \pm2\sqrt{2}=a

so hmmm which is it, the +/-?   well, we know that <span>tan(θ) < 0, that means the tangent of the angle is negative, well, the tangent is opposite/adjacent, the only way that fraction can be negative, is that if either, no both, just either opposite or adjacent is negative.

well, we know the opposite is +1, so, the adjacent then has to be negative, so is -2√(2) then

</span>\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies cos(\theta)=\cfrac{-2\sqrt{2}}{3}
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