# Absolute Value of Trigonometric Function

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## 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.

Although this article appears correct, it's inelegant. There has to be a better way of doing it.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.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 `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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 $90 \degrees$