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? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

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