Trig identities are notoriously difficult to memorize: here’s how to learn them without losing your mind.

Starting from the Pythagorean Theorem and similar triangles, we can find connections between sin, cos, tan and friends (read the article on trig).

Can we go deeper? Maybe we can connect sine with *itself* (sin-ception). In math terms, we’re looking for formulas like this (full cheatsheet):

Instead of memorizing these bad mamma jammas, let’s learn to *draw* the formulas. Euler’s Formula makes it easy.

## Connections In Algebra

In algebra, we study relationships like this:

Working out $17^2$ directly is cumbersome. But we can simplify it to:

In the computer era, sure, we can just crunch $17^2$ directly. The important aspect is realizing that $(a + b)^2$ can be broken into simpler ingredients: $a^2, b^2, a, b$. This is useful in factoring, simplifying equations, and so on.

## Connections In Trig

Let’s turn trig into plain English. What does this mean?

Remembering that sine is “height (as a percentage of max)”, this equation asks: *If we add two angles, what is their total height?*

A quick guess might be to combine the individual heights:

It looks clean, but isn’t quite right. If we keep adding up angles, their height increases until the max (100%), then starts decreasing.

The relationship between angle and height can’t be simple addition.

Now here’s the weird thing: I can draw what the new height should be (*It’s right there!*), but I can’t turn my drawing into an equation.

Or can I?

## Drawing With Euler’s Formula

Euler’s Formula lets us create a circular path using complex numbers:

Crucially, multiplying complex numbers performs a rotation. Aha! We can use Euler’s Formula to draw the rotation we need:

- Start with 1.0, which is at 0 degrees.
- Multiply by $e^{ia}$, which rotates by $a$.
- Multiply by $e^{ib}$, which rotates by $b$.
- Final position = $1.0 \cdot e^{ia} \cdot e^{ib} = e^{i(a+b)}$, or 1.0 at the angle (a+b)

The complex exponential $e^{i(a+b)}$ is pretty gnarly. Just like breaking apart $17^2$, let’s multiply out the pieces:

Now we’re talking! This version easily separates the horizontal position (real component) and vertical position (imaginary component):

- Combined height: $ \sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a) $
- Combined width: $ \cos(a + b) = \cos(a)\cos(b) – \sin(a)\sin(b) $

Boom: two annoying-to-remember trig identities in a single computation. Not a bad deal.

## Understanding The Equation

Now that we’ve found the equation, let’s grok its meaning. When we add the heights, here’s what’s happening:

- The full height of the blue triangle ($\sin(a)$) can’t be used, since the red triangle doesn’t extend as far. (Why? When we add angle $b$, we’re moving at a steeper angle with the same hypotenuse. We gained vertical distance and lost horizontal distance.) We’re effectively “sliding back” $\sin(a)$, reducing it by a factor of $\cos(b)$.
- The full height of the red triangle ($\sin(b)$) can’t be used either, since it’s at an angle. We’re “turning” $\sin(b)$, reducing it by a factor of $\cos(a)$.

Remember that sine and cosine are percentages. In this case,

or

Sure, we would *like* to get the full height of each triangle. But from the diagram, we see $a$ slides back and $b$ is twisted, so height we *actually* get is reduced. Think of each cosine as a tax on your height, reducing the amount you take home. (*Have a height of .90? That’s nice, Papa Cosine will let you keep 75%. Pay up the rest, sucka!*).

Now, what happens for small angles, like $\sin(.01 + .02)?$

We could plug and chug this. But I’m guessing the result is about:

Why? My mental diagram for small angles is this:

There’s no perceptible difference between the ideal heights ($\sin(a)$ and $\sin(b)$) and the “taxed” versions ($\sin(a)\cos(b)$ and $\sin(b)\cos(a)$).

- For tiny angles, $\sin(a + b)$ is a vertical line. It barely loses any height due to the parts sliding or twisting.
- For small angles, cosine (the percent we keep), is close to 100%. We’re keeping the vast, vast majority of the height we have.
- $\sin(x) \sim x$ is a common approximation for small angles (often used in Calculus). Essentially, it says $\sin(x)$ is a line for a brief time period. For small angles, $\sin(a + b) \sim \sin(a) + \sin(b) \sim a + b$.

For cosine, we have a similar diagram:

- This time, the conversion factor matches up (cosine with cosine, sine with sine).
- The full width of the first triangle ($\cos(a)$) gets scaled down to match the width of the second.
- The sine term is negative since it pushes us backwards, reducing our height. We can use similar triangles to extract out this piece.

I’m not typically thinking about the parts in the diagram, though it’s nice to see how they work a few times. If you just need the trig identity, crank through it algebraically with Euler’s Formula.

## Why do we care about trig identities?

Good question. A few reasons:

**1. Because you have to (the worst reason).** Many trig classes have you memorize these identities so you can be quizzed later (argh). You don’t need to *memorize* them, you can work out the formula in about a minute. Save your precious brain space for something else.

**2. We can now “factor” trig functions into simper parts.** We can now separate sine into smaller parts, which is useful in Calculus.

For example, to find the derivative of sine, we need:

and we let $dx$ go to zero. This is tricky to work on directly, but using the $\sin(a + b)$ formula we have

As $dx$ goes to zero, $\cos(dx) = 1$ (zero angle is full width), so we have:

And as $dx$ goes to zero, $\sin(dx)$ and $dx$ become equal:

Plugging this in, we get $\cos(a)$ as the derivative of $\sin(a)$. Phew! Working with trig functions isn’t always easy, but at least it’s manageable.

**3. It’s computationally efficient.** If you’re doing a computer graphics, and frequently calculating sine/cosine (for dot products let’s say), trig identities are useful shortcuts. In the past, these identities were used similar to log tables to make hand-done calculations easier.

**4. Math is about seeing connections.** Because trig functions are derived from circles and exponential functions, they seem to show up everywhere. Sometimes you simplify a scenario by going from trig to exponents, or vice versa.

**5. Deepen your knowledge of Euler’s Formula.** Master Euler’s formula and you’ve mastered circles. And from there, the world! (*Editor’s note: Kalid’s pinky appears to be affixed to his mouth. We’re working on it.*)

See, Euler’s formula lets us *draw* a circle and read off a position. That’s amazing! We can avoid a lot of painful geometry with a few multiplications. If you’re doing any advanced math, letting Leonhard Euler deep into your soul is well worth it. He’s good company.

That’s it for today. Happy math.

## Appendix: Resources and Extended Formulas

You can mix & match trig identities to create a bunch of new ones.

**Subtraction formula: replace b with -b**

**Double-angle formula: replace b with a**

This makes sense: after accounting for the conversion factor, we add the height to itself.

**Half-angle formula: replace and solve**

Start with the double-angle formula and solve for $\sin(a)$, which is half the angle used in $\sin(2a)$. Trig without tears (a great resource and name) has more details:

http://brownmath.com/twt/double.htm

A few other references I found helpful:

- http://www.tc.umn.edu/~nydic001/docs/unpubs/Eulers_Formula_and_Trig_Identities.pdf
- https://www.wyzant.com/resources/lessons/math/trigonometry/eulers-formula-trig-identities
- http://www.ctralie.com/Teaching/Euler/
- http://kunklet.people.cofc.edu/MATH220/euler.pdf

NavneetNovember 4, 2015 at 11:49 amNice article . Thanks.

aishaNovember 6, 2015 at 8:56 amWell done again, superbly done.

It gives beginners like me to think intutively about derivative of tan function

Tan is not bounded by circle but it behaves like sine it does not return it shoots to infinity.as distance to wall is always same whole benefit of increasing angle recieved by wall height (no discount) the percentage of increase in wall height is solely dependent on height of ladder hypotenuse.

aishaNovember 6, 2015 at 9:45 amKindly comment sir, if I am on right track.thank u for such beautiful insight

kalidNovember 6, 2015 at 4:20 pmHi Aisha, glad you enjoyed it! Yep, tan(x) is similar to sin(x), except it grows without bound (until it crosses 90 degrees). I’d like to do a follow-up on the derivatives of trig functions, but you’re on the right track.

Hitoshi YamauchiNovember 12, 2015 at 12:35 pmNice explanation!

Some typos:

We could plug and chug this. But I’m guessing the result is about:

\displaystyle{\sin(.01 + .02) \sim \sin(.03) \sim .03}

instead of

\displaystyle{\sin(.01 + .02) \sim \sin(.03) ~ .03}

The ~ makes just a small space and I think it should be \sim.

Also please search {sin(dx)}, sin(2a), sin(a-b), and sin(a+a). I think these are all “\sin” instead of just “sin”.

Thanks for the great article!

kalidNovember 12, 2015 at 5:09 pm@Hitoshi: Thank you! Just fixed those all up :).

RyannMay 8, 2016 at 8:33 pmI don’t understand the scaling. For instance, why is Sin(B) scaled by Cos(A) ?

kalidMay 12, 2016 at 5:57 amGood question. So, the simple case we might think:

sin(a + b) = sin(a) + sin(b)

That equation, if correct, means we combine the full height of each angle when moving around a circle (for example, a 30 and 20 degree angle have the same height as a 50 degree angle). However, as we move, we are curving, so we don’t really get the full “sin(b)” height. The fraction we actually get [cos(a)] is what we need to scale it down by.

Another way to put it: if we added heights directly, that means adding angles [sin(1), sin(2), sin(3)] would just keep increasing our height. But our height on a circle tops out at 90 degrees, and 91 degrees is actually less tall. Without the cos(a) factor, we can’t reduce our height to account for this.

RyanJune 11, 2016 at 10:30 pmI see. But, why is the scaling factor Cos(A) & not anything else ?

MarcusJune 19, 2016 at 8:54 amThis.

I don’t see how cos(b) should be the ratio we want to scale back sin(a) with?

cakelieAugust 4, 2016 at 5:42 am@Marcus

The hypotenuse of the bottom blue triangle is always 1, as we’re in a unit circle.

The length of the bottom side of the red triangle is between 0 and 1, and is equal to cos(b).

So the red triangle will cover a distance on the hypotenuse of the blue triangle, and that distance will equal cos(b). This is also the percentage of the blue hypotenuse that is covered, as the length is constant at 1.

So if cos(b) = 0.5, exactly 50% of the bottom hypotenuse is covered by the top triangle.

If you reduce one side of a triangle, all other sides of said triangle are reduced by the same ratio. If you make the hypotenuse 50% smaller, the height will be 50% smaller as well. So the actual height becomes sin(a) * 0.5. Or sin(a) * cos(b) in the general case.

In essence, the top triangle is supported by a triangle that is exactly the same as the bottom one but scaled down by a factor of cos(b), which reduces the height by a factor of cos(b).

RohitDecember 10, 2016 at 11:27 pmThere’s a nice diagram illustrating how sin(b) is scaled by cos(b) here: http://math.stackexchange.com/a/1342.

Looking at the diagram on the left and given that cos(a) = adjacent / hypotenuse (where hypotenuse is sin(b)), rearranging the equation gives:

adjacent = cos(a) * hypotenuse = cos(a) * sin(b)

RohitDecember 10, 2016 at 11:29 pmTypo in the above comment, it’s “illustrating how sin(b) is scaled by cos(a)”.

wael . QMay 29, 2016 at 1:52 pmnice math thanks alot , would you help me to prove that ” if ABC is a triangle with AB = AC then 2sinB.cosC = sin A ” using complex numbers >

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LauraJanuary 30, 2018 at 2:58 pmThe diagram under ‘Understanding the Equation’ I really like. Staring at it for a while, you can see how the height of sin(a) transforms to sin(a)cos(b) at the point where the edge of the red triangle touches it. That’s because the hypotenuse of that blue triangle is equal to one, so points along it correspond to percentages of one. Because it’s a straight line, points along it are evenly spaced. Therefore, if you draw horizontal lines from various points along the hypotenuse across to the vertical length, sin, you can find the exact percentage of sin accordingly. Cos (b) is a percentage as Kalid has said before. So if you match that percentage with the height of the blue triangle, the corresponding length is sin (a) cos(b).

I’m having a bit more trouble visualising the sin(b)cos(a) result for the next triangle but hopefully I’ll get there.

LauraJanuary 30, 2018 at 10:06 pmI love the diagram in the section Understanding The Equation. The addition formula for sin (a + b) makes sense looking at the diagram.

The hypotenuse of the blue triangle is one. That makes any point along it measurable as a percentage value. As Kalid has said before, cos and sin are just percentages.

So we can imagine the length of the base of the red triangle of distance cos (b) as a percentage value.

Now the hypotenuse of the triangle with angle (a) can also be considered a way of measuring points along the vertical distance, sin (a). If you, for example, draw a horizontal line connecting the mid way point of the hypotenuse across to the vertical line sin (a), that gives you the distance 50% of sin (a). If you connect a line between the point of distance cos (b) and the vertical line sin (a), that gives you cos (b) of sin(a), or cos(b)sin(a). So we see that the distance sin(a) reduces to cos (b) sin (a) according to this calculation when brought ‘in’ to match up with that point along the hypotenuse.

Now we can do the same trick with calculating the next term in the equation, which as we know will turn out to be cos (a) sin (b).

We know that the red triangle has been tilted up by an angle of (a). That means as Kalid says that we can no longer use the full value of sin (b).

Instead we can mentally construct a new triangle on its side, at ninety degrees from the horizontal, which is attached to the right side of the red triangle. Since it is attached to the red triangle which is askew from the horizontal by angle (a), it is askew from the vertical by that same angle. The hypotenuse of that triangle adjoins the right side of the red triangle. If we were to draw this third triangle in full, the hypotenuse would extend to a distance one and the length of its ‘base’ or vertical distance would be cos (a).

Hence bringing this length ‘in’ to connect with the distance ‘sin (b)’ provides the value cos (a) sin (b).

Reilly BeckstrandApril 30, 2018 at 3:02 pmHey Kalid! I loved the way you used two right triangles to show what was happening. After I realized that sin(a or b) and cos(a or b) could be the side of a triangle it clicked. (for sin(a+b)=sin(a)cos(b)……. that sin(a) is just the vertical distance we “drag” back down the hypotenuse till it meets cos(b). Aha!) I was trying to do the same thing for tan(a+b) and was having a hard time picturing the triangles interacting. Do you know of a good way to visualize that relationship? I know I could just substitute in sin(a+b)/cos(a+b), but that lacks the intuition I’m going for.

kalidMay 1, 2018 at 9:44 amGreat question. I don’t have a visual diagram or intuition for tan(a+b) yet, but I think using the dome-wall-ceiling diagram you might be able to construct it (I’d have to play around a bit). Great idea for a follow-up though.

Apurva SinghDecember 14, 2018 at 12:47 pmv cool blog.. smart way of thinking