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.

Cosine is originally defined as the sine of the complementary 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.

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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