# How to remember the trigonometric identities

I have a test tomorrow and I am having trouble remembering those pesky trigonometrical identities (such as $1-\cos x=2\sin^2(\frac{x}{2})$ )

Do you guys have any tips on how I can remember these?

Thanks :)

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I am flagging this question. We can offer help only on deriving these identities or anything mathematical connected with these identities. – user21436 Dec 21 '11 at 18:55
If $\cos$ is involved, then they usually have only $+$ signs. If $\sin$ is involved, then there is usually a $-$ sign. Other than that, you can memorize $\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and use exponent rules to engineer the needed identity. – alex.jordan Dec 21 '11 at 18:55
@Kannappan I disagree that this should be flagged. Jason is asking for more understanding of the trig identities - an understanding that will make it easier for him to remember them. This falls under the first bulleted item of allowed questions in the faq. – alex.jordan Dec 21 '11 at 19:00
@Kan: "We" - ??? While there are some guidelines regarding what questions are appropriate on this site, maybe you should let us speak for ourselves. – The Chaz 2.0 Dec 21 '11 at 19:11
I have never remembered any identity except that the squares of $\sin$ and $\cos$ add up to $1$, which is just Pythagoras' theorem, and I have never suffered because of that. Why do you want to remember more identities?! – Mariano Suárez-Alvarez Dec 21 '11 at 19:15

For what it's worth:

I just memorize one Pythagorean identity and one of the sum identities. Many of the others (besides the obvious ones: the reciprocal, periodicity, and Pythagorean) can be derived starting with one of the sum formulas.

So, you could just memorize how to derive them. Of course, in a test scenario, this may waste precious time...

Reciprocal identities

The reciprocal identities follow from the definitions of the trigonometric functions.

\eqalign { \sec\theta&= {1\over \cos\theta} \qquad \tan\theta= {\sin\theta\over \cos\theta} \cr \csc\theta&= {1\over \sin\theta} \qquad \cot\theta= {1\over \tan\theta} \cr }

Periodicity relations

The Periodicity relations follow easily by considering the involved angles on the unit circle.

\def\ts{}\eqalign { \sin(\theta)&= \sin(\theta \pm2k\pi) \qquad \csc(\theta)= \csc(\theta \pm2k\pi) \cr \cos(\theta)&= \cos(\theta \pm2k\pi) \qquad \sec(\theta)= \sec(\theta \pm2k\pi)\cr \tan(\theta)&= \tan(\theta \pm k\pi)\phantom{2} \qquad \cot(\theta)= \cot(\theta \pm k \pi) \cr }

\eqalign { \sin(\theta)&= - \sin(\theta -\pi) \qquad \csc(\theta)= - \csc(\theta -\pi) \cr \cos(\theta)&= - \cos(\theta -\pi) \qquad \sec(\theta)= - \sec(\theta -\pi) \cr \tan(\theta)&= - \tan(\theta -\ts{\pi\over2}) \qquad \kern-3pt \cot(\theta)= - \cot(\theta -\ts{\pi\over2}) \cr }

Pythagorean Identities

The first Pythagorean Identity follows from the Pythagorean Theorem (look at the unit circle). The other two Pythagorean Identities follow from the first by dividing both sides by the appropriate expression (divide through by $\sin$ or by $\cos$ to obtain the other two).

\eqalign { \sin^2\theta +\cos^2\theta&=1\cr 1+ \cot^2\theta& =\csc^2\theta\cr \tan^2\theta + 1& = \sec^2\theta}

Sum and difference formulas

Memorize the first sum and difference formula. The second one can be derived from the first using the fact that $\sin$ is an odd function.

One can then derive the last two sum identities by using the first two and the fact that $\cos(\theta-\pi/2)=\sin\theta$.

\eqalign{ \cos(x+y)&=\cos x\cos y-\sin x\sin y\cr \cos(x-y)&=\cos x\cos y+\sin x\sin y\cr \sin(x+y)&=\sin x\cos y+\sin y\cos x\cr \sin(x-y)&=\sin x\cos y-\sin y\cos x\cr }

Double angle formulas

The Double Angle formulas for $\sin$ and $\cos$ are derived by using the Sum and Difference formulas by writing, for example $\cos(2\theta)=\cos(\theta+\theta)$ and using the Pythagorean Identities for the $\cos$ formula (I suppose the formula for $\tan$ should be memorized).

\eqalign{ \sin(2\theta)&=2\sin\theta\cos\theta \cr \tan(2\theta)&= {2\tan \theta\over 1-\tan^2\theta } \cr \cos(2\theta)&= \cos^2\theta-\sin^2\theta \cr &=2\cos^2\theta -1\cr &=1-2\sin^2\theta\cr }

Half angle formulas

The Half-Angle formulas for $\sin$ and $\cos$ are then obtained from the Double Angle formula for $\cos$ by writing, for example, $\cos\theta=\cos(2\cdot{\theta\over2})$

The $\tan$ formula here can easily be obtained from the other two. (Note the forms for the $\cos$ and $\sin$ formulas. These aren't to hard to memorize)

\eqalign{ \cos{\theta\over2}&= \pm\sqrt{1+\cos\theta\over2}\cr \sin{\theta\over2}&= \pm\sqrt{1-\cos\theta\over2}\cr \tan{\theta\over2}&=\pm\sqrt{1-\cos\theta\over1+\cos\theta} }

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For "Periodicity formulas", One may also find it useful to remember ASTC(=All Silver Tea Cups or Add Sugar to Coffee) rule useful to figure out those signs. – user21436 Dec 21 '11 at 19:26
Thanks for helping me :) – Jason Dec 21 '11 at 19:32
I would say that the more useful half-angle identity for the tangent is $$\tan\frac{u}{2}=\frac{\sin\,u}{1+\cos\,u}=\frac{1-\cos\,u}{\sin\,u}$$ – J. M. Dec 22 '11 at 1:38
For that matter, you can get away with only remembering sum of angles formulae if you always remember that sine is odd and cosine is even. – J. M. Dec 22 '11 at 3:31

My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix

$$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

rotates vectors in the plane by an angle $\theta$ and matrix multiplication is the same as composition. Hence, you have identities like

$$\begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

from which it follows

$$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$

and

$$\sin(2\theta) = 2\sin\theta\cos\theta \ .$$

Alternatively, as yoyo says, you could use Euler's identity,

$$e^{i\theta} = \cos\theta + i \sin\theta$$

to find, for instance, that

$$\cos(\theta + \phi) + i\sin(\theta + \phi) = e^{i(\theta + \phi)} = e^{i\theta}e^{i\phi} = (\cos\theta + i\sin\theta) (\cos\phi + i\sin\phi) \ .$$

Hence,

$$\cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi$$

and

$$\sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi \ .$$

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The two approaches are related, of course. The rotation matrix performs rotations in the usual coordinate plane in much the same way that multiplication by $\exp\,i\varphi$ is a rotation in the Argand plane. – J. M. Dec 22 '11 at 3:29
You can use the rotation matrix trick to get the angle addition formulas by using a different variable for each matrix. – Neil G Dec 22 '11 at 9:22
I'm not familiar with matrix multiplication and don't understand how your approach works; would you please expand on it? – James Oct 25 '12 at 16:57
en.wikipedia.org/wiki/Matrix_product – a.r. Oct 25 '12 at 21:25

you should remember the following $$\sin(a\pm b)=\sin(a)\cos(b)\pm\cos(a)\sin(b)$$ $$\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b$$ $$\cos^2x+\sin^2x=1$$ from these (with $x=a=b$) you can get $$\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1$$ and various other rearrangements.

this may not be helpful, but when I couldn't remember sine/cosine of a sum (ie before teaching) i would use $$e^{i\theta}=\cos\theta+i\sin\theta$$ and $$e^{i(\theta+\phi)}=e^{i\theta}e^{i\phi}$$ to find them (or check my memory)

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Thanks for the help :) – Jason Dec 21 '11 at 19:31
And/or $\sin -\theta = -\sin \theta$ while $\cos -\theta = \cos \theta$ to get the additive identities by keeping or dropping the complex term of $e^{i\theta}=\cos\theta+i\sin\theta$ – Ben Jackson Dec 21 '11 at 19:52

Apart from the three sum/difference formulas posted by yoyo which are useful enough to know by heart use the visualization of a moving vector with endpoint on the unit circle. The x-coordinate is cos(a), the y-coordinate is sin(a), a beeing the angle measured counterclockwise from the x-axis.

Most formulas can be derived from this small set of tools fairly easily.

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