Given sin theta = 2/3 and pi/2<theta<pi how do you find the value of other 5 other trigonometric functions?

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Given sin theta = 2/3 and pi/2<theta<pi how do you find the value of other 5 other trigonometric functions?

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Answer 1 · 2015-10-20T00:43:29

$sin (θ) = \frac{2}{3} XXXXXXX csc (θ) = \frac{3}{2}$ $sin(theta) = 2/3color(white)("XXXXXXX")csc(theta)=3/2$
$cos (θ) = - \frac{\sqrt{5}}{3} XXXX sec (θ) = - \frac{3}{\sqrt{5}}$ $cos(theta) = -sqrt(5)/3color(white)("XXXX")sec(theta)=-3/sqrt(5)$
$tan (θ) = - \frac{2}{\sqrt{5}} XXXX cot (θ) = - \frac{\sqrt{5}}{2}$ $tan(theta)=-2/sqrt(5)color(white)("XXXX")cot(theta)=-sqrt(5)/2$

Explanation:

$\frac{π}{2} < θ < π$ $pi/2 < theta < pi$
tells us that $θ$ $theta$ is in Quadrant II

$sin (θ) = \frac{2}{3}$ $sin(theta)=2/3$
tells us that the side ratios are
$XXX$ $color(white)("XXX")$ opposite side: $2$ $2$
$XXX$ $color(white)("XXX")$ hypotenuse: $3$ $3$
and using the Pythagorean Theorem and the fact that we are in Q II
$XXX$ $color(white)("XXX")$ adjacent side: $- \sqrt{5}$ $-sqrt(5)$

By definition:
$sin = \frac{opposite}{hypotenuse} XXXX csc = \frac{hypotenuse}{opposite}$ $sin = ("opposite")/("hypotenuse")color(white)("XXXX")csc=("hypotenuse")/("opposite")$

$cos = \frac{adjacent}{hypotenuse} XXXX sec = \frac{hypotenuse}{adjacent}$ $cos = ("adjacent")/("hypotenuse")color(white)("XXXX")sec=("hypotenuse")/("adjacent")$

$tan = \frac{opposite}{adjacent} XXXX cot = \frac{adjacent}{opposite}$ $tan = ("opposite")/("adjacent")color(white)("XXXX")cot=("adjacent")/("opposite")$

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Given sin theta = 2/3 and pi/2<theta<pi how do you find the value of other 5 other trigonometric functions?

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