How are the 6 basic trigonometric functions related to right triangles?

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How are the 6 basic trigonometric functions related to right triangles?

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Answer 1 · 2014-11-26T05:49:47

The six basic trigonometric functions are:
1. Sine, $sin θ$ $sintheta$
2. Cosine, $cos θ$ $costheta$
3. Tangent, $tan θ$ $tantheta$
4. Cotangent, $cot θ$ $cottheta$
5. Secant, $sec θ$ $sectheta$
6. Cosecant, $csc θ$ $csctheta$

Take the following triangle for example:

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Let the angle marked at A be $θ$ $theta$ .
The longest side of the triangle is the hypotenuse, the side next to the angle is the adjacent and the side opposite to it is the opposite.

Then,

$sin θ = \frac{o p p}{h y p} = \frac{a}{h}$ $sintheta =(opp)/( hyp)=a/h$

$cos θ = \frac{a d j}{h y p} = \frac{b}{h}$ $costheta =(adj)/( hyp)=b/h$

$tan θ = \frac{o p p}{a d j} = \frac{a}{b} = \frac{sin θ}{cos θ}$ $tantheta =(opp)/(adj)=a/b=(sintheta)/(costheta)$

$cot θ = \frac{a d j}{o p p} = \frac{b}{a} = \frac{cos θ}{sin θ} = \frac{1}{tan θ}$ $cottheta =(adj)/(opp)=b/a=(costheta)/(sintheta)=1/tantheta$

$sec θ = \frac{h y p}{a d j} = \frac{h}{b} = \frac{1}{cos θ}$ $sectheta =(hyp)/(adj)=h/b=1/costheta$

$csc θ = \frac{h y p}{o p p} = \frac{h}{a} = \frac{1}{sin θ}$ $csctheta =(hyp)/(opp)=h/a=1/sintheta$

Now if we the triangle were to have sides 3, 4 and 5 units long such that
$a = 4, b = 3, h = 5$ $a=4, b=3, h=5$
The six basic trigonometric functions would be:

$sin θ = \frac{a}{h} = \frac{4}{5}$ $sintheta =a/h=4/5$

$cos θ = \frac{b}{h} = \frac{3}{5}$ $costheta=b/h=3/5$

$tan θ = \frac{a}{b} = \frac{4}{3}$ $tantheta =a/b=4/3$

$cot θ = \frac{b}{a} = \frac{3}{4}$ $cottheta =b/a=3/4$

$sec θ = \frac{h}{b} = \frac{5}{3}$ $sectheta =h/b=5/3$

$csc θ = \frac{h}{a} = \frac{5}{4}$ $csctheta =h/a=5/4$

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How are the 6 basic trigonometric functions related to right triangles?

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