Absolute Value of Trigonometric Function

Theorem

Let $\theta$ be an angle embedded in a Cartesian plane.

Let $\theta$ be such that the vertex of $\theta$ is located at the origin while one arm is coincident with the $x$-axis.

Let $\phi$ be the acute angle made by the other arm with the $x$-axis.

Let $f: \R \to \R$ be a trigonometric function.

Then $\size {\map f \theta}$ is equal to $\map f \phi$.


That is, a trigonometric function of an angle can be calculated by working out its quadrant to determine its sign, then using the equivalent value of what it is between $0$ and $90 \degrees$ to get its value.


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Proof


This theorem requires a proof.
In particular: simple and obvious but I can't be bothered to do a rigorous job, anyone up for it?
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Sources

  • 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Angles larger than $90 \degrees$
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