The Magical Hexagon of Trigonometric Identities

The magical hexagon of trigonometric identities is a handy mnemonic to help you remember a handful of the common trigonometric identities in a very straightforward, beautiful way. It exemplifies patterns between trigonometric formulas that helps one might miss when deriving these formulas algebraically.

Although, a word of caution: I believe mnemonics, while helpful, can be counterproductive in some situations. In the mnemonic “soh-cah-toa”, which stands for “Sine is [the] Opposite [side] over [the] Hypotenuse [of a right triangle], Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent,” through vocalizing the sounds of the mnemonic, we recover the definitions of the sine, cosine, and tangent functions on a right triangle. This is relatively harmless as it does not overshadow any logic underlying these concepts: these definitions/identities for the fundamental trigonometric functions are usually taught to students as is, and memorizing them as such cannot be characterized as taking shortcuts in any way. On the other hand, if one commits the hexagon I am about to present into memory, it covers up a layer of algebraic justification. Consider the following example:

the product of \(\sec(x)\) and \(\cot(x)\) is \(\csc(x)\) because the vertex containing the cotangent function is adjacent to the vertices containing the secant and cotangent functions.

This line of reasoning omits the fact that the product of \(\sec(x)\) and \(\cot(x)\) is \(\cot(x)\) because¹ \[\sec(x) \cdot \cot(x) = \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)} = \csc{x}.\]

The Hexagon

Here is the “Magical Hexagon of Trigonometric Identites.”²

Notice that on the left side we have the functions sine, tangent, and secant. On the right side, parallel with each vertex is its complementary trigonometric function (cofunction): cosine, cotangent, and cosecant. We will now go over all the identities that you can “derive” from the visual mnemonic device with relative ease. Unfortunately, the identities derivable from this trick do not include the angle sum and difference identities, so they will have to be remembered through other means.

Products of Trigonometric Functions

Neighboring Vertices

First of all, if a product of two trigonometric functions, say \(\sin(x)\tan(x)\), corresponded to neighboring vertices on the hexagon, the product expression cannot be “simply” simplified into one function just by rewriting the functions in terms of sine and cosine. In this case, \[\sin(x)\tan(x) = \sin(x) \cdot \frac{\sin(x)}{\cos(x)} = \frac{\sin^2(x)}{\cos(x)},\] which is not equivalent to any one of the six common trigonometric functions.

Sharing an Adjacent Vertex

If two trigonometric functions correspond to vertices for which the shortest path connecting the two vertices is two edges, then they share an unique outer vertex that is adjacent to both of the vertices at hand. Their product is equal to the trigonometric function corresponding to the shared vertex. In the example above, The vertices that \(\tan(x)\) and \(\cos(x)\) corresponds to share the outer vertex \(\sin(x)\) so \[\cos(x)\tan(x) = \cos(x) \cdot \frac{\sin(x)}{\cos(x)} = \sin(x).\]

This pattern also yields the following five formulas in clockwise order on the hexagon.

\[\sin(x)\cot(x) = \sin(x) \cdot \frac{\cos(x)}{\sin(x)} = \cos(x).\] \[\cos(x)\csc(x) = \cos(x) \cdot \frac{1}{\sin(x)} = \frac{\cos(x)}{\csc(x)} = \sin(x).\] \[\sec(x)\cot(x) = \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)} = \csc{x}.\] \[\csc(x)\tan(x) = \frac{1}{\sin(x)} \cdot \frac{\sin(x)}{\cos(x)} = \frac{1}{\cos(x)} = \sec{x}.\] \[\sin(x)\sec(x) = \sin(x) \cdot \frac{1}{\cos(x)} = \frac{\sin(x)}{\cos(x)} = \tan(x).\]

Reciprocals on Diagonally Opposite Vertices

We are probably familiar with the following equations.

\[\cot(x) = \frac{1}{\tan(x)}, \; \sec(x) = \frac{1}{\cos(x)}, \; \csc(x) = \frac{1}{\sin(x)}.\]

In fact, these equations were likely the way these three functions were initially defined and introduced. With a little bit of rewriting, by multiplying out the denominators, we see that

\[\tan(x)\cot(x) = 1, \; \cos(x)\sec(x) = 1, \; \sin(x)\csc(x) = 1.\]

This corresponds to the diagonally opposite vertices on our hexagon. The product of the trigonometric functions corresponding to those vertices diagonally opposite from each other simplify to their midpoint, the center of the hexagon, which is the unit 1. For instance, the example in the picture “demonstrates” that \(\cos(x)\sec(x) = 1\).

Pythagorean Identities

Notice how the three triangles in the hexagon that points downwards are shaded. The rule (which we will explain later) is that the squares of the functions corresponding to the top vertices sum up to the square of the function corresponding to the bottom vertex. For example, following the two light blue arrows on the top downward triangle, the two vertices squared \(\sin^2(x) + \cos^2(x)\) is equal to the bottom vertex squared \(1^2 = 1\).

One of the most important trigonometric identities is the pythagorean identity, which comes in three elementary forms. A quick reminder as to how the trigonometric pythagorean identity is derived: recall the mnemonic “soh-cah-toa” mentioned earlier. Picture a right triangle with an angle \(x\) labelled on one of the other angles. Let the hypotenuse of the triangle be \(h\), let the side adjacent to the angle \(x\) be \(a\), and let the remaining side be \(o\). It should look something like this.

The Pythagorean theorem tells us that the squares of the two legs of the triangle sum up to the square of the hypotenuse. In our diagram, this would be algebraically written as \[o^2 + a^2 = h^2\] Let’s divide the equation by \(h^2\) \[\frac{o^2}{h^2} + \frac{a^2}{h^2} = \left(\frac oh\right)^2 + \left(\frac ah\right)^2 = \frac{h^2}{h^2} = 1.\] Now, recalling from “soh-cah-toa” that \(\sin(x) = \frac oh\) and \(\cos(x) = \frac ah\), we have that \[\sin^2(x) + \cos^2(x) = 1,\] which is the Pythagorean identity. This corresponds to the downward triangle with light blue arrows on the hexagon.

The other two versions of the Pythagorean identity can be derived from the first with ease. We will just have to divide the equation by \(\sin^2(x)\) or \(\cos^2(x)\). Dividing by \(\cos^2(x)\), \[\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)},\] and simplifying algebraically, \[\left(\frac{\sin(x)}{\cos(x)}\right)^2 + 1 = \left(\frac{1}{\cos(x)}\right)^2,\] and thus \[\tan^2(x) + 1 = \sec^2(x).\] This corresponds to the downward triangle with lime/green arrows on the hexagon.

Similarly, Dividing by \(\sin^2(x)\), \[\frac{\sin^2(x)}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)},\] and simplifying algebraically, \[1 + \left(\frac{\cos(x)}{\sin(x)}\right)^2 = \left(\frac{1}{\sin(x)}\right)^2,\] and thus \[1 + \cot^2(x) = \csc^2(x).\] This corresponds to the downward triangle with red/pink arrows on the hexagon.

Cofunction Identities

The nomenclature definitiong these complementary trigonometric functions imply some sort of relationship between them and the original trigonometric functions. Specifically, in a right triangle, if one trigonometric function on one of the other angles evaluate to a certain angle, the complementary trigonometric function on the other angle should evaluate to the same value. If we call the first angle \(x\), the other angle, also known as the complementary angle, is \(\frac{\pi}{2} - x\).³ According to the way the trigonometric functions were placed on the vertices of the hexagon, functions are parallel (same height) to their complementary counterparts. Therefore, horizontal lines connect functions with this complementary relationship.⁴

On the top row, we see that \[\sin(x) = \cos\left(\frac{\pi}{2} - x\right)\] we can prove this using the cosine angle difference identity, noting that \(\sin(\frac{\pi}{2}) = 1\), and \(\cos(\frac{\pi}{2}) = 0\). \[\begin{align*} \cos\left(\frac{\pi}{2} - x\right) & = \cos\left(\frac{\pi}{2}\right)\cos(x) - \sin\left(\frac{\pi}{2}\right)\sin(x) \\ & = 0\cdot \cos(x) + 1\cdot \sin(x) = \sin(x). \end{align*}\]

It is also pretty clear that the arrows go in the opposite direction as well (even though I did not label them), since \[\sin\left(\frac{\pi}{2} - x\right) = \cos\left(\frac{\pi}{2} - \left(\frac{\pi}{2} - x\right)\right) = \cos(x).\]

With this property demonstrated for sine and cosine, the complementary property for the remaining trigonometric functions can be easily verified.

\[\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\cos\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right)} = \cot\left(\frac{\pi}{2} - x\right).\] And so \(\tan(\frac{\pi}{2} - x) = \cot(x)\).⁵

\[\sec(x) = \frac{1}{\cos(x)} = \frac{1}{\sin\left(\frac{\pi}{2} - x\right)} = \csc\left(\frac{\pi}{2} - x\right).\] And so \(\sec(\frac{\pi}{2} - x) = \csc(x)\).

Conclusion

We have covered and derived the trigonometric identities that can be derived from the hexagon. This could be a great tool for precalculus and calculus students who struggle to recall (or forgot due to lack of use) trigonometric identities quickly. Furthermore, due to the visual nature of the mnemonic, you will not have to compute trigonometric expressions such as \(\sec(x)\cot(x)\) in your head by expanding them into fractions and cancelling out the cosine term. Instead, visualizing the hexagon could help simplify these mental computations too! The magical hexagon is a elegant, useful, and fun tool, as long as you know how its results are supposed to be derived, which is why this post covered so many derivations of trigonometric identities in depth.

1. And then we could talk about the definitions of the cotangent, secant, and cosecant functions as reciprocals of tangent, cosine, and sine. ↩

2. The original LaTeX used to produce the graphics for the magical hexagon were originally created by Joseph Nilsen on TeXample.net and can be found here. ↩

3. Yes, I know, radians are far superior. ↩

4. This is one of the cases where the diagram is not that useful. But nevertheless, as long as I can intuitively incorporate a trigonometric identity into the diagram, I will. ↩

5. Apply the same argument that was used to demonstrate \(\sin(\frac{\pi}{2} - x) = \cos(x)\) above. ↩