Show that tangent function is tangent to unit circle

[Mondo]

Hello,

I stumbled upon this graphics which shows various trigonometric functions on a unit circle.



I wonder how could we show that tangent function [imath]tan = \frac{sin(\theta)}{cos(\theta)}[/imath] is really the hypotenuse of the big triangle OAE?

Thanks

 

[BigBeachBanana]

Hello,

I stumbled upon this graphics which shows various trigonometric functions on a unit circle.

View attachment 33305

I wonder how could we show that tangent function [imath]tan = \frac{sin(\theta)}{cos(\theta)}[/imath] is really the hypotenuse of the big triangle OAE?

Thanks

[imath]\tan(\theta)[/imath] is not the hypotenuse, but the opposite side of the triangle OAE.
[math]\cos(\theta)=\frac{adj}{hyp}=\frac{OA}{OE}=\frac{1}{OE} \implies OE=\frac{1}{\cos(\theta)}=\sec(\theta)\\ \sin(\theta)=\frac{opp}{hyp}=\frac{AE}{OE}=\frac{AE}{\sec(\theta)}\implies AE = \sin(\theta)\sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)[/math]

 

[pka]

Hello,

I stumbled upon this graphics which shows various trigonometric functions on a unit circle.

View attachment 33305

I wonder how could we show that tangent function [imath]tan = \frac{sin(\theta)}{cos(\theta)}[/imath] is really the hypotenuse of the big triangle OAE?

Look at this plot.
Do you still say that the tangent function is tangent to to the unit circle?

 

[Mondo]

Thank you @BigBeachBanana that's it.

tan(θ) is not the hypotenuse, but the opposite side of the triangle OAE.

Right my mistake.

Look at this plot.
Do you still say that the tangent function is tangent to to the unit circle?

@pka heh you show a graph of tangent function with some ellipse fit into it. This is not what I meant. My question was about a unit circle with cos and sin functions well defined and now prove is needed that tangent function is really the segment shown on the graph from first post.
Anyway, thank you for showing an alternative view on the question

 

[BigBeachBanana]

Thank you @BigBeachBanana that's it.

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