How do you find the remaining trigonometric functions of $θ$ $theta$ given $cot θ = \frac{m}{n}$ $cottheta=m/n$ where m and n are both positive?

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $cot θ = \frac{m}{n}$ $cottheta=m/n$ where m and n are both positive?

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Answer 1 · 2017-04-02T16:37:17

$tan θ = \frac{n}{m}$ $tantheta=n/m$

$sin θ = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{n}{\sqrt{n^{2} + m^{2}}}$ $sintheta=(opposite)/(hypoten use)=n/(sqrt(n^2+m^2$

$cos θ = \frac{a d j a c e n t}{h y p o t e n u s e} = \frac{m}{\sqrt{n^{2} + m^{2}}}$ $costheta=(adjacent)/(hypoten use)=m/(sqrt(n^2+m^2)$

$csc θ = \frac{1}{sin θ} = \frac{h y p o t e n u s e}{o p p o s i t e} = \frac{\sqrt{n^{2} + m^{2}}}{n}$ $csctheta=1/sintheta=(hypoten use)/(opposite)=(sqrt(n^2+m^2))/n$

$sec = \frac{1}{cos θ} = \frac{h y p o t e n u s e}{a d j a c e n t} = \frac{\sqrt{n^{2} + m^{2}}}{m}$ $sec=1/costheta=(hypoten use)/(adjacent)=sqrt(n^2+m^2)/m$

Explanation:

First let's identify the ones that are left.

$tan θ, sin θ, cos θ, sec θ, csc θ$ $tantheta, sintheta,costheta,sectheta,csctheta$

But $cot θ = \frac{1}{tan θ} = \frac{a d j a c e n t}{o p p o s i t e} = \frac{m}{n}$ $cottheta=1/tantheta=(adjacent)/(opposite)=m/n$

$tan θ = \frac{o p p o s i t e}{a d j a c e n t} = \frac{n}{m}$ $tantheta=(opposite)/(adjacent)=n/m$

Using the pythagorean theorem

$h y p o t e n u s e^{2} = o p p o s i t e^{2} + a d j a c e n t^{2}$ $hypoten use^2=opposite^2+adjacent^2$

$h y p o t e n u s e = \sqrt{o p p o s i t e^{2} + a d j a c e n t^{2}}$ $hypoten use=sqrt(opposite^2+adjacent^2)$

$h y p o t e n u s e = \sqrt{n^{2} + m^{2}}$ $hypoten use=sqrt(n^2+m^2)$

$sin θ = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{n}{\sqrt{n^{2} + m^{2}}}$ $sintheta=(opposite)/(hypoten use)=n/(sqrt(n^2+m^2$

$cos θ = \frac{a d j a c e n t}{h y p o t e n u s e} = \frac{m}{\sqrt{n^{2} + m^{2}}}$ $costheta=(adjacent)/(hypoten use)=m/(sqrt(n^2+m^2)$

$csc θ = \frac{1}{sin θ} = \frac{h y p o t e n u s e}{o p p o s i t e} = \frac{\sqrt{n^{2} + m^{2}}}{n}$ $csctheta=1/sintheta=(hypoten use)/(opposite)=(sqrt(n^2+m^2))/n$

$sec = \frac{1}{cos θ} = \frac{h y p o t e n u s e}{a d j a c e n t} = \frac{\sqrt{n^{2} + m^{2}}}{m}$ $sec=1/costheta=(hypoten use)/(adjacent)=sqrt(n^2+m^2)/m$

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $cot θ = \frac{m}{n}$ $cottheta=m/n$ where m and n are both positive?

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Science

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... and beyond

How do you find the remaining trigonometric functions of θtheta given cotθ=mncottheta=m/n where m and n are both positive?

1 Answer

Explanation:

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $cot θ = \frac{m}{n}$ $cottheta=m/n$ where m and n are both positive?