# Is there a more efficient method of trig mastery than rote memorization?

I would like to get alot better at trig than I am. What is the best/most efficient method?

Thanks much in advance

Joe

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What sorts of trig are you trying to learn? The unit circle? Identities? –  mixedmath Sep 18 '11 at 18:34
In the P.S. of Arturo's answer here are some basic tricks one needs to know in order to learn (and avoid memorizing to many of) the trig identities (I thought there was a much more extensive version of that somewhere, but I can't seem to find it right now). The other answers there might also help. –  t.b. Sep 18 '11 at 18:40
If you can ever get to learning about complex numbers and Euler's formula, I'd say that those definitely ease the task of deriving trigonometric identities from scratch. –  Guess who it is. Sep 18 '11 at 23:13
I think solving more and more problems on based on trig manipulation is the best way to master them,however it works for any other formulas. –  VelvetThunder Sep 18 '11 at 23:13
–  Jesse Madnick Sep 19 '11 at 5:08

I would emphasize how to derive trigonometric identities from a few ones. Learn:

• how to derive the relations between the direct functions of the same angle from the definition of the trigonometric functions and the Pythagorean formula;
• maxima, minima, zeroes and period of each function;
• if a function is odd or even;
• the trigonometric functions of 0º, 30º, 45º, 60º and 90º;
• the relations between functions of symmetric, complementary and supplementary angles;
• the relations between functions of angles whose difference is 180º;
• the relations between functions of angles whose sum is 360º;
• the inverse trigonometric functions;
• the addition formulas of sin and cos;
• how to derive the subtraction formulas of sin and cos;
• how to derive the addition and subtraction formulas of tan and cot;
• how to derive the double and half angle formulas;
• how to derive the sum to product formulas;
• how to solve some elementary trigonometric equations;
• (triangle) sin and cos laws;
• Heron's formula;
• derivatives of direct and inverse trigonometric functions.

$$\sin (\alpha +\beta )=\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin \beta ,\tag{A}$$

if we set $\alpha =\beta =a$, we get

$$\sin 2a=2\sin a\cdot \cos a.\tag{1}$$

And from

$$\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta \tag{B}$$

for $\alpha =\beta =a$, we have

$$\cos 2a=\cos ^{2}a-\sin ^{2}a.\tag{2}$$

Using the Pythagorean identity

$$\cos ^{2}a+\sin ^{2}a=1,\tag{C}$$

if $\cos a\neq 0$, then $$\begin{eqnarray*} \sin 2a &=&2\sin a\cdot \cos a=2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a+\sin ^{2}a} \\ &=&\dfrac{2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin ^{2}a }{\cos ^{2}a}}=\dfrac{2\dfrac{\sin a}{\cos a}}{1+\dfrac{\sin ^{2}a}{\cos ^{2}a}} \\ &=&\dfrac{2\tan a}{1+\tan ^{2}a}. \end{eqnarray*}\tag{3}$$

Similarly $$\begin{eqnarray*} \cos 2a &=&\cos ^{2}a-\sin ^{2}a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a+\sin ^{2}a} \\ &=&\dfrac{\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin ^{2}a}{\cos ^{2}a}}=\dfrac{1-\dfrac{\sin ^{2}a}{\cos ^{2}a}}{1+\dfrac{\sin ^{2}a }{\cos ^{2}a}} \\ &=&\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}. \end{eqnarray*}\tag{4}$$

Then

$$\tan 2a=\dfrac{\sin 2a}{\cos 2a}=\dfrac{\dfrac{2\tan a}{1+\tan ^{2}a}}{\dfrac{ 1-\tan ^{2}a}{1+\tan ^{2}a}}=\dfrac{2\tan a}{1-\tan ^{2}a}.\tag{5}$$

Added 2. The linear equation in $\sin x$ and $\cos x$

$$A\sin x+B\cos x=C\tag{6}$$ can be solved by a resolvent quadratic equation in $\tan \frac{x}{2}$, by writting the $\sin x$ and the $\cos x$ functions in terms of $\tan \frac{x}{2 }$ (set $x=2a$ in $(3)$ and $(4)$):

$$\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}},\tag{7}$$

$$\cos x=\dfrac{1-\tan ^{2}\dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}.\tag{9}$$

The equation $(6)$ is equivalent to $$\begin{eqnarray*} A\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}+B\frac{1-\tan ^{2}\dfrac{x}{ 2}}{1+\tan ^{2}\dfrac{x}{2}} &=&C, \\ 2A\tan \dfrac{x}{2}+B-B\tan ^{2}\dfrac{x}{2} &=&C+C\tan ^{2}\dfrac{x}{2}, \\ \left( B+C\right) \tan ^{2}\dfrac{x}{2}-2A\tan \dfrac{x}{2}+C-B &=&0. \end{eqnarray*}\tag{10}$$

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• Some arrays of trigonometric identities have rhythmic patterns that can help you remember them: \begin{align} \sin(x+y) & = \sin x\cos y + \cos x\sin y \\ \sin(x-y) & = \sin x\cos y - \cos x\sin y \\ \cos(x+y) & = \cos x\cos y - \sin x\sin y \\ \cos(x-y) & = \cos x\cos y + \sin x\sin y \end{align}

• But you should not simply learn these four identities SEPARATELY. If you know the first identity, you immediately get the second one from it by knowing that the sine function is odd and the cosine is even, and in the same way you get the fourth from the third. And you can get the sine identities from the cosine identities and vice-versa by recalling that $\sin x = \cos(\pi/2 - x)$ and $\cos x = \sin(\pi/2-x)$. If you don't know how to do things like these, you're missing something you should learn.

If you know how to find $\sin(x+y)$ as a function of the sines and cosines of $x$ and $y$, that immediately tells you the double-angle formula $\sin(2x)=2\sin x\cos x$, so you should not just SEPARATELY memorize that. Again, if you don't know that, then learn it. Similarly for the other double-angle formulas.

If you know the four identities above, that tells you how to show that $$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x\tan y}$$ by first recalling that $\tan = \sin/\cos$ and then dividing the numerator and denominator both by $\cos x\cos y$. So again, don't merely SEPARATELY learn the identity above; learn how it emerges from the others.

And this applies to trigonometric identities generally.

• Learn how $\sin^2 x + \cos^2 x = 1$ emerges from the Pythagorean theorem. Draw a triangle whose hypotenuse length is $1$. Then the opposite side is $\sin x$ and the adjacent side is $\cos x$. Similarly if you draw a triangle whose adjacent side has length $1$, then the opposite side is $\tan x$ and the hypotenuse is $\sec x$, so we get another Pythagorean identity: $1+\tan^2 x = \sec^2 x$.

• Another useful thing to do if you can, is to tutor those who are first learning the subject who are not as adept as you are. After taking a dozen-or-so such students through the whole course, you'll find some things much more firmly fixed in place in your mind.

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"And you can get the sine identity from the cosine by recalling that $\sin x = \cos(\pi/2 - x) and \cos(x) = \sin(\pi/2-x)$." Heh; for a long time I used to derive those from the sum formulas, rather than the sum formulas from that. Later, of course, I did $\cos(x) = \sin(x+\pi/2)$ by thinking about the graphs... –  Arturo Magidin Sep 18 '11 at 23:15
And you can get $\sec^2 x=\tan^2 x+1$ and $csc^2 x=\cot^2 x+1$ from the more familiar $\sin^2 x+\cos^2 x=1$ by dividing by $\cos^2 x$ and $\sin^2 x$, respectively. And $\sin^2 x+\cos^2 x=1$ and $\cos 2x=\cos^2 x-\sin^2 x$ together yield $\cos 2x=2\cos^2 x-1$ and $\cos 2x=1-2\sin^2 x$, from which the half-angle formulas $\cos^2 x=(1+\cos 2x)/2$ and $\sin^2 x=(1-\cos 2x)/2$ are immediate. As Michael says, in the long run it’s easier to remember relationships amongst the identities than to remember them as discrete facts. –  Brian M. Scott Sep 19 '11 at 0:11
If all else fails, remember that $$e^{i \theta}=\cos(\theta)+i\sin(\theta)$$ and that $$e^{i(\theta + \phi)}=e^{i \theta}e^{i \phi}.$$ All trig identities may be derived from this. Thank Euler! –  Giuseppe Negro Sep 19 '11 at 0:36
It's a bit perverse, but before I learned Euler's formula, what I used to remember were the double angle formulae; a look at their form allowed me to "derive" the more general addition formulae ("derive" in the sense that since e.g. $\sin\,2x=2\sin\,x\cos\,x$, I knew that the addition formula's two terms had to have both sines and cosines, and similarly for $\cos\,2x$). –  Guess who it is. Sep 19 '11 at 1:21
@J.M. If I were you, I will call your trick a mnemonic. Sounds more respectable that way. :-) –  Srivatsan Sep 19 '11 at 16:01

The complex-valued function $h(\theta)=\cos(\theta)+i \sin (\theta)$ is uniquely determined by the following three properties:

1. $h$ is continuous;
2. $h(\theta+\phi)=h(\theta)h(\phi)$;
3. $\lvert h(\theta)\rvert=1$, $h(2\pi)=1$.

This means that any other property of $\cos$ and $\sin$, including all trigonometric formulas, can be derived from the above.

Those properties are especially easy to remember if you write $e^{i \theta}=\cos(\theta)+i\sin(\theta)$ (this great idea is due to Euler), and this notation provides also a convenient algebraic way of deriving trig identities.

Example. We want to expand $\cos^2(\theta)$. Write $\left( e^{i\theta}\right)^2= e^{i 2 \theta}$. The left hand side is

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

while the right hand side is

$$\cos 2\theta + i \sin 2 \theta.$$

Equating real parts we get

$$\cos^2\theta=\sin^2\theta+\cos 2\theta,$$

and since $\lvert e^{i \theta}\rvert^2=1$, that is $\cos^2\theta+\sin^2\theta=1$, we conclude

$$\cos^2\theta=\frac{1+\cos 2 \theta}{2}.$$

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And if you are into calculus, use the addition (or subtraction) formulae to get the derivatives of sin, cos, and tan. See what additional fact is needed to get the derivatives.

And if you are metaphysical, ponder why the addition formula for tan involves only tan, while the addition formulae for sin and cos involve both sin and cos.

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