Would this be a valid way of proving a trigonometric value despite one of the sides of the triangle being negative?

Question

I want to prove that this is correct but is using triangles with negative lengths a valid way to do so?

https://i.stack.imgur.com/2ASGY.jpg

Answer 1

No it is not valid, length can never be negative. You can however use other was to prove your question.

since $\cos\theta $ is negative it lies in either the $2^{nd}$ or $3^{rd}$ quadrant , but the value of $\tan(\theta)$ is positive so the angle $\theta$ must be in the $3^{rd}$ quadrant.

Doing the same thing as you did before but with hypotenuse $3$ gives the result.

Note: You could also use the fact that;

$\tan(\theta) = \dfrac{\sqrt{1-\cos^2(\theta)}}{-\cos(\theta)} = \dfrac{\sqrt{1-\frac19}}{\frac13}= \sqrt8 =2\sqrt2 $

Answer 2

To use algebraic lengths (i.e. that can take a negative sign), you need to be explicit about the orientation conventions, and specify the positive directions of the lines as well as of the rotation angles.

Usually this is done in the trigonometric circle with a standard coordinate frame, and the oblique lines (hypothenuses/radii) are considered positive, not algebraic.