Trigonometry

Reciprocal identities

${\sin x = \frac{1}{\csc x}} \text{\ \ \ \ \ } {\csc x = \frac{1}{\sin x}}$
${\cos x = \frac{1}{\sec x}} \text{\ \ \ \ \ } {\sec x = \frac{1}{\cos x}}$

Secant and cosecant not taught in Sweden.

${\tan x = \frac{1}{\cot x}} \text{\ \ \ \ \ } {\cot x = \frac{1}{\tan x}}$

If you prefer expressing them in sine and cosine:

${\tan x = \frac{\sin x}{\cos x}} \text{\ \ \ \ \ } {\cot x = \frac{\cos x}{\sin x}}$

Interestingly, note that tan is 1/cotan, but sine is not 1/cosine.

Offset identities

$\sin \theta = \cos(\frac{\pi}{2} - \theta)$

Imagine the unit circle. Instead of starting at zero, you start at pi/2. Then, due to subtraction, you walk clockwise, not counterclockwise.

Unit circle

Inverse trig functions

There are different definitions of the domain of inverse trig functions. Multi-valued arcsin is a thing. Mathematica in particular might define the cotangent differently from Sweden. But here are the rules for basic-uni Swedish math.

$\arcsin x: -\pi/2 \le x \le \pi/2, -1 \le y \le 1$

TODO Explain how to think, don't just make a reference table

Law of sines

We can use it to find stuff about nonright triangles.

Angle A, opposing side a;

$\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$

Num and denom can be swapped.

Law of cosines

Can tell you stuff about nonright triangles. Like with the Law of Sines, the capital letters denote angle opposing a side denoted in lower case.

$c^2 = a^2 + b^2 - 2ab cosC$

The Pythagorean Theorem is a special case of this Law.

Pythagorean identity

In Swedish, it's called trigonometriska ettan. cos²x + sin²x = 1²

Should be obvious if you look at the unit circle and remember the Pythagorean Theorem.

Half-angle formula

Double-angle formula

sin(2x) = 2 sin(x) cos(x)

Proof?

Example: Find the sine of π/12

Using the double angle formula, we know that

sin(π/6) = 2 sin(π/12) cos(π/12).

We also know from our standard angles that sin(π/6) is 1/2. So

$1/2 = 2 \sin(\pi/12) \cos(\pi/12)$ $\iff \frac{1}{4\sin(\pi/12)} = \cos(\pi/12)$

The reason I left cosine by itself is that we'll exploit the Pythagorean identity. First, let's square this thing.

$\frac{1}{16\sin^2(\pi/12)} = \cos^2(\pi/12)$

Now, look at a different equation, the Pythagorean identity:

$\sin^2(\pi/12) + \cos^2(\pi/12) = 1$

We have an expression for $\cos^2(\pi/12)$ , so replace it.

$\sin^2(\pi/12) + \frac{1}{16\sin^2(\pi/12)} = 1$

Multiply the equation so we're rid of the fraction.

$16\sin^4(\pi/12) + 1 = 16\sin^2(\pi/12)$

Now, we can do variable substitution. Say $y := \sin^2(\pi/12)$ .

$16y^2 + 1 = 16y$ $\iff 16y^2 - 16y + 1 = 0$

Solve like any second-degree equation.

$y = \frac{2 \pm \sqrt{3}}{4}$

So, given that we made y the squared sine of π/12, we square root the equation and get our answer.

$\sin(\pi/12) = \sqrt{\frac{2 \pm \sqrt{3}}{4}}$

To figure out whether the plus or minus applies, test them. The result $1/2 + \frac{sqrt{3}}{4}$ is definitely larger than sin(π/12) ought to be, so the plus-minus sign should be a minus.

Euler's formula🔗

General form: e^iθ = cos(θ) + isin(θ)

Insert π and check the value. It gives you zero sin, -1 cos. So: e^iπ = -1. You quickly realize that inputting any multiple of pi will give you a real number.

Trig angle-addition formulas can be derived: substitute (a + b) for θ. Convert to cis(a + b) form. Perform the complex multiplication. The terms with i will correspond to the angle addition formula for sin, since the sine axis in the unit circle and the imaginary axis are both vertical. The terms without i then, of course, correspond to the angle addition formula for cos.

De Moivre's formula follows from Euler's formula. (cosθ + isinθ)ⁿ is (e^iθ)ⁿ, which by simple exponent rules is e^iθn, which by definition is cos(nθ) + isin(nθ).

De Moivre's formula🔗

(cosθ + isinθ)ⁿ is equal to cos(nθ) + isin(nθ). You can state the cisθ in terms of Euler's formula to prove it.

Angle addition

To get the formula for tan, merely do the angle addition for sin, divided by angle addition for cos. Tan is sin/cos.

$\frac{\sin(x) \cos(y) + \cos(x) \sin(y)}{\cos(x) \cos(y) - \sin(x) \sin(y)}$

Multiply both numerator and denominator by $\frac{1}{\cos(x) \cos(y)}$ and you will get a simplified form.

$\frac{\tan(x) + \tan(y)}{1 - \tan(x) \tan(y)}$

What links here

Math