Suppose I have a trig function $T: \Bbb{R} \rightarrow \Bbb{R}$. I want to be able to derive four basic properties:

$$T(x) \cdot T(y)$$ $$T(x) + T(y)$$ $$T(x+y)$$ $$T(cx)$$

where $c$ is some scalar.

I know there are a bunch of identities: reciprocal, quotient, Pythagorean, co-function, even-odd. And then some formulas: product-to-sum, sum-to-product, sum-difference, double angle, half-angle/power-reducing. Here is a list. That's a lot to memorize and some of them seem to overlap.

Why do the four basic properties I mentioned require so many identities to learn? It is a bit cumbersome. Is there an easier way to learn how to do the basic arithmetic operations with trig functions?

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    $\begingroup$ The definitions of $\tan$, $\csc$, $\sec$, and $\cot$ must be memorized (nemonic: each pair of reciprocals has one "co" e.g., $(\sin,\csc)$ are a pair) I memorize $\sin^2(x)+\cos^2(x)=1$ and the angle sum/difference formulas. Everything else can usually be derived from those. $\endgroup$ Jul 7, 2015 at 19:32
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    $\begingroup$ I do the same as Michael Burr. I've memorized the addition angle formulae for sine and cosine and derive other identities from those. $\endgroup$
    – Mark Viola
    Jul 7, 2015 at 19:51
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    $\begingroup$ I think that this question is relevant. $\endgroup$
    – Ivo Terek
    Jul 7, 2015 at 19:56
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    $\begingroup$ The $e^{ix}e^{iy}=e^{i(x+y)}$ thing is useful. (If you haven't seen this before, just treat "$e^{ix}$" as shorthand for "$\cos(x)+i\sin(x)$." Where $i^2=-1$. Note: $e^{ix}=\cos(x)+i\sin(x)$ is actually a theorem, rather than just notation, but it takes a while to prove.) $\endgroup$ Jul 8, 2015 at 7:22
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    $\begingroup$ The way to learn (memorize, if you prefer) the formulas is to do 1,000 exercises that call for them. You will not only learn the formulas, you'll learn which ones are useful in which situations. $\endgroup$ Jul 8, 2015 at 9:32

5 Answers 5


I always found recalling $e^{ix}=\cos x+i\sin x$ useful for quickly deriving the sum of angles formulae, e.g.



$$e^{ix+iy}=e^{ix}e^{iy}=(\cos x+i\sin x)(\cos y+i\sin y).\tag{2}$$

Expanding (2), equating with (1) and separating real and imaginary parts gives you the formulae.

You can then get the double angle formulae easily.

Wait, there's more!

We have $(e^{ix})^n=(\cos x+i\sin x)^n$.

But we also have $$(e^{ix})^n=e^{inx}=\cos(nx)+i\sin(nx),$$ so we get

$$(\cos x+i\sin x)^n = \cos(nx)+i\sin(nx).$$

For example, consider $n=2$, then expanding gives:

$$\cos^2 x-\sin^2 x = \cos(2x)$$ and $$2\sin x\cos x=\sin(2x).$$

This is another way to get the double angle formulae, but you can get more trig identities by letting $n=3, 4, \ldots$. In general, for positive integer $n$ we have

$$\cos(nx) = \Re\left((\cos x+i\sin x)^n\right) =\Re\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin ^k(x)$$ and $$\sin(nx) = \Im\left((\cos x+i\sin x)^n\right)=\Im\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin^k(x).$$

Expanding and simplifying will give you nice trig identities. This is called De Moivre's Theorem.

  • $\begingroup$ This is very very cool. I didn't anticipate learning this. It also makes it very easy to keep track of the sign changes as you take derivitives of the basic trig functions. $\endgroup$ Jul 8, 2015 at 18:46
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    $\begingroup$ @StanShunpike This is a special case of series bisection / multisection. $\endgroup$ Jul 8, 2015 at 21:09
  • $\begingroup$ @BillDubuque Wow! I've wondered what you call that for a while. Awesome! $\endgroup$ Jul 8, 2015 at 23:43

I find that four suffice. $$\cos^2 (x) + \sin^2(x) = 1 \tag{1}$$ $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) \tag{2}$$ $$\sin(a+b)=\sin(a)cos(b)+\sin(b)\cos(a) \tag{3}$$ $$\text{trig}(x) = \text{cotrig}(\frac{\pi}{2}-x) \tag{4}$$

The Pythagorean identity $(1)$ is easy to manipulate. Divide through by $cos^2(x)$ alternatively by $sin^2(x)$ to find the other forms

$$1 + \tan^2(x) = sec^2(x) \tag{5}$$ $$\cot^2(x) +1 = csc^2(x) \tag{6}$$

For the angle addition formulas $(2)$ and $(3)$, we can apply odd and even identities to quickly derive the angle subtraction identities:

$$\cos(a+\color{red}{(-b)})=\cos(a)\cos(\color{red}{(-b)})-\sin(a)\sin(\color{red}{(-b)})$$ $$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \tag{7}$$


$$\sin(a+\color{red}{(-b)})=\sin(a)cos(\color{red}{(-b)})+\sin(\color{red}{(-b)})\cos(a)$$ $$\sin(a-b)=\sin(a)cos(b)-\sin(b)\cos(a) \tag{8}$$

We can also let $a=b$ and substitute in $(2)$ and $(3)$ to find double angle forms

$$\cos(a+a)=\cos(a)\cos(a)-\sin(a)\sin(a)$$ $$\cos(2a)=\cos^2(a)-\sin^2(a) \tag{9}$$

$$\sin(a+a)=\sin(a)cos(a)+\sin(a)\cos(a)$$ $$\sin(2a)=2\sin(a)cos(a) \tag{10}$$

Combining $(9)$ with the Pythagorean identity $(1)$ gives two more.

$$\cos(2a)=1-2\sin^2(a) \tag{11}$$ $$\cos(2a)=2\cos^2(a)-1 \tag{12}$$

If you want a half angle formula, you may as well let $u=2a$ in the previous four identities. Just mind your squares and roots. What happening to tangent? Divide any related $sin$ by $cos$ to get what you need.

Let's not forget product to sum identities! If we take $(2)$ and $(7)$ and add the equation, we find

$$\cos(a+b) + \cos(a-b)=2\cos(a)\cos(b)$$ $$\cos(a)\cos(b) = \frac12 (\cos(a+b) + \cos(a-b)) \tag{13}$$

Similar combinations will give the remaining product to sum identities.

As for $(4)$, $trig(x) = cotrig(\frac{\pi}{2}-x)$, I'm referring to cofunction identities, which all have the same form. For example, $\sin(x) = \cos(\frac{\pi}{2}-x).$ That's essentially six more identities.

We have over twenty identities at our disposal now, including the few that I've mentioned but neglected to outright derive.

  • $\begingroup$ This was super helpful $\endgroup$ Jun 29, 2017 at 9:50

From the comments to the question:

Michael Burr:

The definitions of $\tan$, $\csc$, $\sec$, and $\cot$ must be memorized (mnemonic: each pair of reciprocals has one "co" e.g., $(\sin,\csc)$ are a pair) I memorize $\sin^2(x)+\cos^2(x)=1$ and the angle sum/difference formulas. Everything else can usually be derived from those.

This is great because it simplifies everything down to a core set of principles. The problem is then memorizing the sum/difference formulas. I found the following mnemonic in another answer


Sine, Cosine, Sign, Cosine, Sine

Cosine, Cosine, Co-Sign, Sine, Sine

The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $\alpha, \beta, \alpha, \beta$ in each line), and know that "Sign" means to use the same sign as in the compound argument ("+" for angle sum, "-" for angle difference), while "Co-Sign" means to use the opposite sign.

Together, I think this makes a pretty good game plan for tackling trig. For a little more fun, see Antinous' awesome answer.


Stan's answer will probably do a better job than mine; while running the risk of repeating information: memorize $\sin^2(x)$ $+$ $\cos^2(x)$ $=$ $1$, dividing through by $\cos^2(x)$ will yield the identity $\tan^2(x)$ + $1$ $=$ $\sec^2(x)$.

Additionally, dividing through by $\sin^2(x)$ will yield $1 +$ $\cot^2(x)$ = $\csc^2(x)$. Furthermore, intuitively understanding the graphs of $\sin(x)$ and $\cos(x)$ will make life easier.

Finally, going through the unit circle derivations of many trigonometric identities will prove useful in being able to "unstick" yourself should you get stuck.

Edit: Of course, also knowing the basics: $\tan(x)$ $=$ $\frac{\sin(x)}{\cos(x)}$, $\csc(x)$ $=$ $\frac 1{sin(x)}$, $\sec(x)$ $=$ $\frac 1{cos(x)}$ is fundamental.


A systematic way to implement what pbs wrote, is mentioned in the book A = B in section 1.5:

Let $w = \exp(ix)$, then $\cos x = (w +w^{-1})/2$ and $\sin x = (w - w^{-1}/(2i$). So equality of rational expressions in trigonometric functions can be reduced to equality of polynomial expressions in w. (Exercise: Prove, in this way, that $\sin 2x = 2\sin x\cos x$.)


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