**This post shows how the core trigonometric definitions, relations and addition theorems can be simply and intuitively visualized.**

In the above diagram, the 3 classic trigonometric identities are now obvious:

$$ \sin^2 \theta + \cos^2 \theta = 1; \quad \quad 1+\tan^2 \theta = \sec^2 \theta; \quad \quad 1+\cot^2 \theta = \csc^2 \theta $$

Note that most of us learnt that the *definitions* of $\sec, \csc, \cot$ were $1/\cos, 1/\sin, 1/\tan$, respectively. However, from a historical and geometrical point of view, this is not quite true.

**Cos**ine is originally defined as the **sine** of the **co**mplementary angle, $90-\theta$.

Tan is equal to the length of the **tangent** resulting from a radial line at angle $\theta$, and $\sec$ is equal to the length of the **secant** derived from $\theta$.

And from this, we directly have $\cot$ which is therefore the tangent of the complementary angle, and $\csc$ which is the secant of the complementary angle.

Here is a common alternative to the above version.

In this version all those on the lower-right hand side of the radial line (sin, tan, sec) are associated with the angle $\theta$.

And all those on the upper-left hand side of the radial line (cosine, cotan, cosec) are associated with the complementary angle $90-\theta$.

Note that in both of them you learn that the tan function is intimately related to the tangent of a circle, and not just the ratio “opposite over hypotenuse’.

**Hyperbolic Trig Functions**

Unfortunately there is not a single diagram that as succinctly shows the analogous hyperbolic relationships and identities

From this diagram, we can clearly see $ \cosh^2 \theta – \sinh^2 \theta = 1 $, but the other two hyperbolic trig identities are not so obvious.

Figure 3. Hyperbolic Functions. This presents a direct analogy of figure 2.

See here for some other good versions.

**Addition Theorems**

Looking at the vertical sides, we have: $\sin(\alpha+\beta) = \sin \alpha \; \cos \beta + \cos \alpha \; \sin \beta$.

And comparing the horizontal sides, we have: $\cos(\alpha+\beta) = \cos \alpha \; \cos \beta – \sin \alpha \; \sin \beta$.

The white triangle on the left-hand side shows: $$\tan (\alpha + \beta) = \frac{\tan \alpha +\tan \beta}{1-\tan \alpha \tan \beta} $$.

The following diagram assists in understanding why

$\cos 2 \alpha = \cos^2 \alpha – \sin^2 \alpha $$

and

$\sin^2 \alpha = 2 \sin \alpha \; \cos \alpha.$

My name is Martin Roberts. I have a PhD in theoretical physics. I love maths and computing. I’m open to new opportunities – consulting, contract or full-time – so let’s have a chat on how we can work together!

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My other contact details can be foundhere.

# Other Posts you may like

- A Simple Formula for Sequences and Series
- Lissajous Curves
- The Perimeter of an Ellipse
- Multiple Pendulums

## 3 Comments

## Manfred Weis

A very nice collection of mathematical gems that are hard to find when you need them. I really enjoyed every single page; hope you keep the spirit up.

## Martin Roberts

Thank you so much for these encouraging words. Sometimes I find it hard to keep writing as I don’t feel it comes naturally to me, so your kind words are very much appreciated. 🙂

## Lachlan Karslake

Thank you Martin for this post, I very much appreciate it. As someone who struggles with Trigonometry due to shaky foundations this exactly what I needed. Cheers.