Notes on a review of trigonometry

These are notes taken during RPI's MATH 1010 course, relating to a review of trigonometry for calculus.

It is recommended to use Ctrl F or the equivalent search function to find the relevant section, as these notes are quite long.

Trigonometric functions

Trigonometric functions are functions of angles. An angle is positive when measured counterclockwise from the initial ray, and negative when measured counterclockwise from the initial ray. Calculus typically uses radians, where $2\pi$ radians is equal to 360 degrees.

Angles in degrees can be converted to radians by multiplying by:

$$ \frac{\pi}{180} $$

Similarly, angles in radians can be converted to degrees by multiplying by:

$$ \frac{180}{\pi} $$

Trigonometric relations

$$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$

$$ \csc \theta = \frac{1}{\sin \theta} $$

$$ \sec \theta = \frac{1}{\cos \theta} $$

$$ \cot \theta = \frac{1}{\tan \theta} $$

The trig song for remembering trig functions

Mr. Darnbrook, wherever you are - countless students will thank you for this song you taught me 5 years ago...

One, two three!

Three, two, one!

Two under the bar!

Square root all that's not one!

Square root of three over three!

One, square root of three!

Now that you know the song!

You can sing it again, to me!

And this is how you get to this:

Explanation:

"One, Two, Three" is the top of the first row of the table

"Three, Two, One" is the top of the second row

"Two under the bar" means to make 2 the denominator of every number (so 1 will become $\frac{1}{2}$, 2 will become $\frac{2}{2}$, you get the idea)

"Square root all that's not one" means to add a square root to every numerator (in rows 1) and 2) that, as you might have guessed, is not 1)

"Square root of three over three" is for the first column of the third row ($\tan 30^\circ$)

"One, square root of three" is for the second and third columns of the third row ($\tan 45^\circ$ and $\tan 60^\circ$)

And you've got that done!

Signs of the trig functions

The trig functions obey:

You can remember this with the acronym All Students Take Calculus. For (positive) trig angles greater than $2\pi$, we can find the quadrant of the angle by subtracting the greatest multiple of $2\pi$ - for negative, do the same but add the greatest multiple of $2\pi$.

Reference angles

For all angles in the form:

$$ \theta = \frac{k\pi}{n} $$

The reference angle is:

$$ \frac{\pi}{n} $$

For instance, the reference angle of $\frac{5\pi}{6}$ is $\frac{\pi}{6}$.

Simplifying angles

For any angle that is greater than $\pm 2\pi$:

$$ \sin \theta = \sin(\theta \pm 2n\pi) $$

$$ \cos \theta = \cos(\theta \pm 2n\pi) $$

$$ \tan \theta = \tan(\theta \pm 2n\pi) $$

where $n$ is any positive integer. Therefore, $\sin\left(\frac{9\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$, and $\cos \left(-\frac{11\pi}{3}\right) = \cos \left(-\frac{5\pi}{3}\right)$. This can be used to simplify a very large angle into a much smaller one that can be more easy to work with.

Finding the trig values of an arbitrary angle

Let's say we want to find the value of the trig functions at $\theta = -\frac{3\pi}{4}$.

First, we know that $-\frac{3\pi}{4}$ is an integer multiple of $\frac{\pi}{4}$, so we know that the reference angle is $\frac{\pi}{4}$.

We can then use the trig song to find the values of the trig functions at $\frac{\pi}{4}$:

$$ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

$$ \tan \frac{\pi}{4} = 1 $$

But remember that $-\frac{3\pi}{4}$ is in the 3rd quadrant, where (using "all students take calculus") only tangent is positive. So we apply the correct signs (-sin, -cos, +tan):

$$ \sin \left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \cos \left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \cos \left(-\frac{3\pi}{4}\right) = +1 = 1 $$

Trig identities

The Pythagorean identities are easiest to remember. The first pythagorean identity is:

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

If we divide both sides by $\cos^2 \theta$ we get the second pythagorean identity:

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

Finally, if we add "co" to the second identity (tangent -> "co" tangent, secant -> "co" secant), we get the third:

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

Finally, the half-angle and double-angle formulas are useful:

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

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

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

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

Solving trig equations

A solution $\theta$ of $\sin \theta = k$ satisfies the fact that if you know one solution $\theta_1$ of the equation, then:

$$ \theta = \theta_1 \pm 2\pi n $$

Similarly, a solution of $\theta_1$ of $\cos \theta = k$ satisfies the fact that if you know one solution $\theta_1$ of the equation, then:

$$ \theta = \theta_1 \pm 2\pi n $$

This is because sine and cosine have infinite solutions that are distanced $2\pi$ apart. The same is true for tangent - there are infinite solutions that are distanced $\pi$ apart:

$$ \theta = \theta_1 \pm \pi n $$

For example, if we were to solve $\cos \theta = \frac{1}{2}$, we'd first use the trig song/table to find the values that satisfy that equation. In our case, from the trig song/table, it is clear that:

$$ \theta_1 = \frac{\pi}{3} $$

satisfies the equation. However, it is not the only value that satisfies the equation. To find the other values that satisfy the equation in $[0, 2\pi]$, we consider all the integer multiples of our first solution $\theta$ that are within the interval. That is, we have:

$$ \theta = \theta, 2\theta, 3\theta, \dots n\theta $$

In our case, the integer multiple angles of our solution $\theta_1$ within $[0, 2\pi]$ are:

$$ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{6\pi}{3} $$

Now, from these angles, we choose the angles that lie in a quadrant that yield the same sign as $\theta_1$ for the given trig function in the problem. For example, if $\theta_1$ was positive, then we'd choose the angles that lie in quadrants 1 and 4, as those quadrants are where cosine (the trig function in the problem) is positive. Likewise, if $\theta_1$ was negative, then we'd choose the integer multiples that lie in quadrants 2 and 3, as those quadrants are where cosine is negative. Adjust this based on the type of trig function used (sine/cosine/tangent) and the sign of $\theta_1$ (+ or -). E.g. if $\theta_1$ is a solution of the tangent function and negative, then we'd choose solutions in quadrants 2 and 4.

For our case, $\theta_1$ is positive and a solution to the cosine function, so we choose only the angles that lie in quadrants 1 and 4. Therefore, the solutions in $[0, 2\pi]$ are:

$$ \theta = \frac{\pi}{3}, \frac{5\pi}{3} $$

Since there are infinitely many solutions in general, however, it would be more correct to say that:

$$ \theta = \frac{\pi}{3} \pm 2\pi n, \frac{5\pi}{3} \pm 2\pi n $$

Inverse functions

An inverse function is a function that undoes the effect of a function. Given a function $f$ and an inverse function $f^{-1}(x)$, then:

$$ f^{-1}(y) = x $$

$$ f(x) = y $$

Note that the "raised to -1" does not mean $\frac{1}{f}$. It is a notation to denote the inverse operation. An inverse function can be found by swapping $x$ and $y$ in a function, and then solving for $y$. The value of $y$ is the inverse function. An inverse function can only be found if a function is a function (passes vertical line test) and is one-to-one (passes horizontal line test). Graphically $f^{-1}$ is a reflection of $f$ across $y = x$.

Inverse trigonometric functions

To find the inverse functions of the trigonometric functions, we must restrict their domain so that on that interval, the trig function is one-to-one. For example, we restrict $\sin(\theta)$ to $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos(\theta)$ to $\theta \in [0, \pi]$, and $\tan (\theta) = [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Evaluating inverse trigonometric functions

Unlike typical trigonometric functions, inverse trigonometric functions only have one possible output for any given input as inverse trig functions are one-to-one. Say we were to evaluate $\theta = \cos^{-1}(\frac{1}{2})$. To solve this, we swap $\frac{1}{2}$ and $\theta$, then change the inverse cosine to regular cosine, to get:

$$ \cos \theta = \frac{1}{2}, \theta \in [0, \pi] $$

Evidently, the answer is that $\theta = \frac{\pi}{3}$. You may wonder where we got the bounds from. The bounds for the output of each inverse trigonometric function are (you should memorize this):

So the format of evaluating inverse trig functions is to swap the $x$ and $\theta$, then set $\theta \in \text{[your bounds]}$, then solve.

When taking the input of an inverse trig function of a trig function of a value, in the form:

$$ x = \sin^{-1}(\sin(\theta)) $$

$$ x = \cos^{-1}(\cos(\theta)) $$

$$ x = \tan^{-1}(\tan(\theta)) $$

If $\theta$ is within the range of the inverse trig function, $x = \theta$; otherwise, it has to be manually computed.