## How I Remember Trig Identities Part 1

There are many trigonometric identities that students are expected to know for math olympiad, and it’s not always easy to remember all of them. This post will walk you through how I remember just a few key identities and use them to derive the rest. (Of course it is best to memorise them until they become second nature, but even the best of us have mental blocks from time to time.)

The full list of identities that I will be covering can be found at this link.

## 1. Definition of trigonometric functions

For any angle $\theta$, we can draw a right-angled triangle as above. The 3 sides of the triangle have standard names. The longest side is the hypotenuse (which you can remember from Pythagoras’ Theorem). The other 2 sides are called adjacent and opposite because, well, they are adjacent to and opposite from the angle itself.

There are 3 basic trigonometric ratios: sine, cosine and tangent. In school I was taught to remember these ratios using the mnemonic

“Toa Cah Soh”

which, in Hokkien, means “fat-legged auntie“.  Not exactly an image I would like to have in my head, but it definitely sticks. Each of the words reminds me of what the ratios are: for example, “toa” tells me that tangent = opposite / hypotenuse.

The other 3 trig functions which appear often are secant, cosecant and cotangent. Cotangent is easy to remember since it stands out. A common mistake is thinking that secant = 1 / sine and cosecant = 1 / cosine. To remember this, I remind myself that reciprocals “switch” things around, and so

$\sec = \displaystyle\frac{1}{\cos}, \hspace{1em} \csc = \displaystyle\frac{1}{\sin}, \hspace{1em} \cot = \displaystyle\frac{1}{\cot}.$

## 2. Is it positive or negative?

Ever wonder how to remember which trig functions are positive in which quadrant? The following diagram is helpful:

S for sine, T for tangent, C for cosine, A for all. I remember it (as I was taught in school) as SATC, some others remember it as ACTS.

Another way to remember it is by remembering that the hypotenuse is always a positive value, while whether the opposite/adjacent side is positive of negative depends on its coordinate. For example, in the triangle below:

The adjacent side is negative while the opposite and hypotenuse are positive. Hence, any function which involves the adjacent side will be negative. Only the sine (and cosecant) function doesn’t involve the adjacent side, hence it is the only trig function that is positive.

This diagram can also help us remember the even-odd identities. For any angle $\theta$, the angle $-\theta$ is the reflection in the x-axis. From the SATC diagram, we can tell that for the sine and tangent functions, flipping in the x-axis will reverse the sign of the result while the sign is maintained for the cosine function. Thus we obtain

\begin{aligned} \sin (-\theta) &= - \sin\theta &&- (1) \\ \tan(-\theta) &= - \tan\theta &&- (2) \\ \cos (-\theta) &= \cos \theta &&- (3) \end{aligned}

## 3. Co-Function Identities

The angles $\theta$ and $\frac{\pi}{2} - \theta$ are intimately linked because they are complementary angles (i.e. add up to 90°). In terms of trig functions, they share the same hypotenuse and have their adjacent & opposite sides switched. Hence, from this triangle we can see that

\begin{aligned} \sin (\frac{\pi}{2} -\theta) &= \cos\theta &&- (4) \\ \cos(\frac{\pi}{2}-\theta) &= \sin\theta &&- (5) \\ \tan (\frac{\pi}{2}-\theta) &= \cot \theta &&- (6) \end{aligned}

(The analysis for angles that are not acute is a bit more complicated but uses the same concept.)

## 4. Pythagorean Identities

I would really suggest memorising

$\sin^2 \theta + \cos^2 \theta = 1 - (7)$,

even though one could derive it by dividing the Pythagorean identity $(\text{opp})^2 + (\text{adj})^2 = (\text{hyp})^2$ throughout by $(\text{hyp})^2$.

To derive the other 2 Pythagorean identities, divide identity $(7)$ by a term that would make one of the 2 terms on the LHS equal to 1. For example, dividing by $\sin^2 \theta$ yields

\begin{aligned} 1 + \frac{\cos^2 \theta}{\sin^2 \theta} &= \frac{1}{\sin^2 \theta}, \\ 1 + \cot^2 \theta &= \csc^2 \theta - (8). \end{aligned}

Similarly, dividing $(7)$ by $\cos^2 \theta$ yields

$1 + \tan^2 \theta = \sec^2 \theta - (9).$

This is it for today’s edition. We’ll go through the rest of the identities tomorrow!