If $csc θ = \frac{4}{3}$ $csc theta=4/3$ , what is the sin, cos, tan, sec, and cot?

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If $csc θ = \frac{4}{3}$ $csc theta=4/3$ , what is the sin, cos, tan, sec, and cot?

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Answer 1 · 2018-03-27T15:43:28

See below.

Explanation:

Instead of using formulas, it'd be easier to solve it geometrically, with a right triangle.

enter image source here

Since $csc θ = \frac{1}{sin θ} = \frac{hypotenuse}{opposite} = \frac{c}{a} = \frac{4}{3}$ $csc theta = 1/sintheta = "hypotenuse"/"opposite"=c/a = 4/3$ , this means that $a$ $a$ and $c$ $c$ are multiples of $3$ $3$ and $4$ $4$ , respectively.

In other words, we have $c = 4 k$ $c=4k$ and $a = 3 k$ $a=3k$ , for a real number $k$ $k$ .
By the Pythagorean theorem, $b = \sqrt{c^{2} - a^{2}} = \sqrt{16 k^{2} - 9 k^{2}} = \sqrt{7} \cdot k$ $b = sqrt(c^2-a^2) = sqrt(16k^2-9k^2) = sqrt(7)*k$ .

Finally, for trigonometric functions :

$sin θ = \frac{opposite}{hypotenuse} = \frac{a}{c} = \frac{3}{4}$ $sin theta = "opposite"/"hypotenuse" = a/c = 3/4$
$cos θ = \frac{adjacent}{hypotenuse} = \frac{b}{c} = \frac{\sqrt{7}}{4}$ $cos theta = "adjacent"/"hypotenuse" = b/c = sqrt7/4$

$tan θ = \frac{opposite}{adjacent} = \frac{a}{b} = \frac{3}{\sqrt{7}}$ $tan theta = "opposite"/"adjacent" = a/b = 3/sqrt7$
$cot θ = \frac{1}{tan θ} = \frac{b}{a} = \frac{\sqrt{7}}{3}$ $cot theta = 1/tan theta = b/a = sqrt7/3$

$sec θ = \frac{hypotenuse}{adjacent} = \frac{c}{b} = \frac{4}{\sqrt{7}}$ $sec theta = "hypotenuse"/"adjacent" = c/b = 4/sqrt7$ .

Answer 2 · 2018-04-02T11:46:05

As below.

Explanation:

$csc θ = \frac{4}{3}$ $csc theta = 4/3$

![https://hononegah.learning.powerschool.com/hhearn/2014-2015honorspre-calculus/cms_page/view/16304893]( useruploads.socratic.org )

$sin θ = \frac{1}{csc θ} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$ $sin theta = 1/csc theta = 1 / (4/3) = 3/4$

${cos}^{2} θ = 1 - {sin}^{2} θ = 1 - \frac{9}{16} = \frac{7}{16}$ $cos^2 theta = 1 - sin^2 theta = 1 - 9 / 16 = 7 / 16$

$cos θ = \pm \frac{\sqrt{7}}{4}$ $cos theta = +- sqrt7 / 4$

$sec θ = \frac{1}{cos θ} = \pm \frac{4}{\sqrt{7}}$ $sec theta = 1 / cos theta = +- 4 / sqrt7$

$tan θ = \frac{sin θ}{cos θ} = \pm \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \pm \frac{3}{\sqrt{7}}$ $tan theta = sin theta / cos theta = +- (3/4) / (sqrt7 / 4) = +-3 / sqrt7$

$cot θ = \frac{1}{tan θ} = \pm \frac{\sqrt{7}}{3}$ $cot theta = 1 / tan theta = +- sqrt7 / 3$

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