This post contains the info I pretty much reiterate verbatim for every trigonometry (or calculus, etc.) student I've tutored. There will be a future post on how to use this one.

## Definitions

These should be the only things you have to accept without reason.

Sine and Cosine: Given the unit circle <!(a circle of radius 1 centered at the origin)> and a point on it making an angle $$\theta$$ counter-clockwise from the $$+x$$-axis, its coordinates are $$(\cos\theta,\sin\theta)$$. More explanation to come in another post.

Tangent: When we instead want the slope of the line from the origin to the point, we use $$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$$.

Secant, Cosecant, Cotangent: When we stumble upon the reciprocals of those trig functions, we nickname them, where
\begin{align} \csc\theta & = \frac1{\sin\theta} & \sec\theta &= \frac1{\cos\theta} & \cot\theta &= \frac1{\tan\theta} \end{align}

I typically remember the weird bits of these definitions with the saying ("Math is backwards."!Math is backwards.)(math-backwards)

(start math-backwards)

This is to remember that even though "sine" and "secant" sound first compared to "cosine" and "cosecant", their definitions apply to second things: the $$y$$-coordinate and the reciprocal of cosine. The same saying can be used for the angle being measured counter-clockwise and for the slope having $$y$$ over $$x$$.

(stop math-backwards)

Also, it's always good to focus on sine and cosine when applying the definitions. (How?)(how-focus)

(start how-focus)

Use the definitions of the other four functions to always change them into sines and cosines, [at least in your head.][(When I read "$$\sec t$$" I say in my head "one over cosine t".)] That way, you basically have to memorize fewer formulas, and get to see what's happening more. ([For instance,][then there's only one Pythagorean Identity.])

(stop how-focus)

## Pythagorean

By the definition of sine and cosine, they form the legs of a right triangle with hypotenuse $$1$$, the radius of the unit circle. Using the Pythagorean theorem, we get the identity

$\sin^2\theta + \cos^2\theta = 1$

Note that writing "$$\sin^2\theta$$" is a weird convention, and (deserves explanation.!Explanation.)(sine-squared)

(start sine-squared)

It means the same as $$(\sin \theta)^2$$ (evaluate the sine, then square) but takes less effort to write and [contrasts better with $$\sin(\theta^2)$$ when spoken.][(since one is said "sine squared theta" while the other is said "sine theta squared")]

(stop sine-squared)

Also note that this formula relates to the [equation of the unit circle.][($$x^2 + y^2 = 1$$)]

If you want this formula in terms of the other trig functions, then divide both sides either by $$\sin^2\theta$$ or $$\cos^2\theta$$ to get

$1 + \cot^2\theta = \csc^2\theta \\ \tan^2\theta + 1 = \sec^2\theta$

## Even-Odd

Using the definition of sine and cosine, notice how the coordinates of the point change when looking at a negative angle (going clockwise instead). Since the horizontal position doesn't care whether you're going clockwise or not, cosine is even, that is,

$\cos(-\theta) = \cos\theta$

Since the vertical position swaps sign when the angle does, sine is odd, that is,

$\sin(-\theta) = -\sin\theta$

Plugging those results into the definitions for the other trig functions shows that sine, cosecant, tangent, and cotangent all must be odd, while cosine and secant are even.

If you're wondering why the terms "even" and "odd" are being used here, I have a post about that.

## Sum-Difference

For these, we want to know how taking a trig function of a sum or difference of angles depends on functions of the angles separately. We start with the sum formulas for sine and cosine, which have many possible derivations, though most students prefer to just memorize them. They are

$\sin(u+v) = \sin u \cos v + \cos u \sin v \\ \cos(u+v) = \cos u \cos v - \sin u \sin v$

To memorize them, I recommend practicing (plugging in special cases.!Special cases.)(sum-diff-cases)

(start sum-diff-cases)

Notice that each formula's right side has a $$\cos v$$ term, followed by a $$\sin v$$ term. Since the formulas must be true for every $$v$$, they must be true when $$v=0$$. In that case, the right side [simplifies to whatever is in front of $$\cos v$$,][(since $$(\cos v, \sin v) = (1, 0)$$)] while the left side [simplifies to either $$\sin u$$ or $$\cos u$$.][(since $$u+v=u$$)] This tells you which term should have which functions of $$u,$$ while the plus and minus signs can either be memorized or derived from the $$v=-u$$ case.

(stop sum-diff-cases)

To get the tangent formula, divide the sine equation by the cosine one, and clean up the ugly fraction on the right side by dividing both numerator and denominator by both $$\cos u$$ and $$\cos v$$, so that everything becomes tangents. (Steps?)(sum-diff-tan)

(start sum-diff-tan)

\begin{aligned} \tan(u+v) &= \frac{\sin(u+v)}{\cos(u+v)} \\ &= \frac{\sin u\cos v + \cos u\sin v}{\cos u\cos v - \sin u\sin v} \\ &= \frac{\sin u\cos v + \cos u\sin v}{\cos u\cos v - \sin u\sin v}\cdot\frac{1/(\cos u\cos v)}{1/(\cos u\cos v)} \\ &= \frac{\tan u + \tan v}{1-\tan u\tan v} \end{aligned}

(stop sum-diff-tan)

The result is

$\tan(u+v) = \frac{\tan u + \tan v}{1-\tan u\tan v}$

To get the difference formulas, replace $$v$$ with $$-v$$ in each equation, and apply the even-odd rules to find the effects on each right-hand side. (It swaps the sign of every written addition and subtraction symbol.)

Note: Co-Function Identities.
Some classes have students memorize the specific case of $$u=\frac\pi 2$$, with results such as $$\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$$. They focus on this case because it illustrates that sine and cosine are shifted, reversed versions of each other, as are the co- versions of each trig function. However insightful these formulas may be, I find them easily derivable and rarely useful, so I don't include them here.

## Double Angle

These are a special case of the sum formulas, where $$u=v$$. I'll swap to using $$\theta$$ to highlight that they have only one variable.

$\sin(2\theta)=2\sin \theta\cos \theta \\ \cos(2\theta)=\cos^2 \theta - \sin^2 \theta \\ \tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$

The cosine formula may remind you of the Pythagorean identity, and applying that identity to the right side gives two other forms:

\begin{aligned} \cos^2 \theta - \sin^2 \theta &= 2\cos^2 \theta - 1 \\ &= 1 - 2\sin^2 \theta \end{aligned}

## Power-Reducing (and Half Angle)

The benefit of the alternate forms of the cosine double angle formula is that each contains just a sine or cosine on the right. If you solve for them, you get the power-reducing formulas, named after how they have a square of a trig function expressed in terms of $$\cos(2\theta)$$. They are

$\sin^2 \theta = \frac12(1-\cos(2\theta)) \\ \cos^2 \theta = \frac12(1+\cos(2\theta))$

And as with the sum-difference formulas, the tangent is obtained by dividing the other two to get

$\tan^2 \theta = \frac{1-\cos(2\theta)}{1+\cos(2\theta)}$

To get the half angle formulas, replace $$\theta$$ with $$\frac12\theta$$ in each equation.

## Product-to-Sum

Now we're going to swap back to $$u$$'s and $$v$$'s, which means we're going back the sum-difference formula again. There's some cute linear algebra you can use here, but you can get along fine without it.

Our objective is to express a product of trig functions in terms of a sum of trig functions. So, take a look at the [sum and difference formulas for sine,][(the ones for $$\sin(u+v)$$ and $$\sin(u-v)$$)] and focus on the product term $$\sin u\cos v$$ that doesn't change sign. If you were to add the two equations together, it'd cancel the other term, leaving you with two of this one, easy to solve for and get

$\sin u\cos v = \frac12\big(\sin(u+v) + \sin(u-v)\big)$

If you subtract the equations instead, you get

$\cos u\sin v = \frac12\big(\sin(u+v) - \sin(u-v)\big)$

(You could also have gotten that one by swapping the places of $$u$$ and $$v$$, applying the odd rule for sine on the right.)

Doing the same thing for the cosine sum and difference formulas gives you

$\cos u\cos v = \frac12\big(\cos(u+v) + \cos(u-v)\big) \\ \sin u\sin v = -\frac12\big(\cos(u+v) - \cos(u-v)\big)$

## Sum-to-Product

You may notice that we are really close to being able to go [the other direction.][(expressing sums like $$\sin u + \cos v$$ in terms of products like $$\sin u\cos v$$)] All we have to do is multiply everything by $$2$$ (or by $$-2$$ for the last one) and make new variables for $$u+v$$ and $$u-v$$, which I'll call $$s$$ and $$t$$, and we get the sum-to-product formulas

$\sin s + \sin t = 2\sin\left(\tfrac12(s+t)\right) \cos\left(\tfrac12(s-t)\right) \\ \sin s - \sin t = 2\cos\left(\tfrac12(s+t)\right) \sin\left(\tfrac12(s-t)\right) \\ \cos s + \cos t = 2\cos\left(\tfrac12(s+t)\right) \cos\left(\tfrac12(s-t)\right) \\ \cos s - \cos t = -2\sin\left(\tfrac12(s+t)\right) \sin\left(\tfrac12(s-t)\right)$

(Once again, that second formula can be derived from the first with $$v$$ replaced with $$-v$$ and the odd rule for sine applied on the left.)

(If you're wondering about that $$\frac12(s\pm t)$$ that showed up, it's the same linear algebra as before, just for $$u$$, $$v$$ and $$s$$, $$t$$, instead of $$\sin u\cos u$$, $$\cos u\sin v$$ and $$\sin(u+v)$$, $$\sin(u-v)$$.)