8.6 Trigonometric Identities
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- simplify trigonometric expressions
- know and use the fundamental Pythagorean identity, and derive the others from it
- know and use the even/odd, sum, difference, and double angle identities
- legally manipulate expressions involving trig functions
In this section, we're going to learn ways to legally manipulate expressions involving trig functions. Beyond using definitions and fraction simplifying techniques, we're also going to learn some essential trig identities. These are more like...... secret identities. They tell you what trig expressions are equivalent—secretly the same guy, like Clark Kent and Superman (spoiler alert). Applying these true statements to rearrange expressions into more useful forms is really just like a puzzle! Gear up for the idea that sometimes you just need to throw a lot of spaghetti at the wall and see what sticks, and that's okay. Just a) write down anything you know to be true, and b) don't be afraid to try something and see what happens. Okay, let's go.
Technique #1 when simplifying or manipulating expressions featuring trig function is simple to write everything in terms of sines and cosines and see if anything magical happens. Technically, in doing this, we are using facts we already learned, which I'll now tell you are known as the Reciprocal Identities,
\[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{1}{\tan x}, \notag \]
and the Quotient Identities,
\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x}. \notag\]
Simplify the expression \( \csc x \tan x + \sec x \).
Solution
Let's translate everybody to sines and cosines only:
\[ \csc x \tan x + \sec x = \frac{ 1}{\sin x} \cdot \frac{ \sin x}{\cos x} + \frac{1}{\cos x} \notag \]
Ope, nice, we have matching terms on top and bottom that we can cancel! NOTE: The quantity we're cancelling is \( (\sin x)\) AS A WHOLE. You cannot cancel like this: \( \frac{\sin x}{x} \rightarrow \sin \). That's a crime—a sin, if you will. Anyways, let's simplify:
\[ \frac{ 1}{\textcolor{red}{\sin x}} \cdot \frac{ \textcolor{red}{\sin x}}{\cos x} + \frac{1}{\cos x} = \frac{1}{\cos x} + \frac{1}{\cos x} = \frac{2}{\cos x} \quad \text{ or } \quad 2 \sec x \notag \]
Simplify the expression \( \sin x \cot x - \frac{1}{\sec x} \).
- Answer
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0.
Pythagorean Identities
When we look at triangles on the unit circle to determine values of \( \sin \theta\) and \(\cos \theta\), we always have the hypotenuse of length \(1\) (since it's a radius of the circle). Then the vertical leg of the right triangle is of size "\( \sin \theta\)" (whatever that number turns out to be) and the horizontal leg is of size "\( \cos \theta\)."
Writing down the Pythagorean Theorem for this triangle gives us our first important identity. Several more follow as a direct consequence of it as well.
For any consistent expression \(A\), we have
\[ \sin^2 A + \cos^2 A = 1. \notag \]
(Notation note: \( \sin^2 A \) just means \( (\sin A)^2\).)
We also have
\[ \tan^2 A + 1 = \sec^2 A, \quad \quad 1 + \cot^2 A = \csc^2 A \notag \]
You don't need to memorize the other two, as long as you know the first one. We can derive them from the OG using a strategic division across both sides of the equation. For example, divide by \( \cos^2 A\) and see what happens.
Solution
We divide through the whole equation by \( \cos^2 A\).
\[ \frac{1}{\cos^2 A} \cdot ( \sin^2 A + \cos^2 A ) = 1 \cdot \frac{1}{\cos^2 A} \quad \rightarrow \quad \frac{ \sin^2 A}{\cos^2 A} + 1 = \frac{1}{\cos^2 A} \notag \]
Now we notice that \( \frac{ \sin^2 A}{\cos^2 A} = \left( \frac{\sin A}{\cos A} \right)^2 = \tan^2 A \), and note that \( \frac{1}{\cos^2 A} = \sec^2 A\), by definition of secant. All together, we get
\[ \tan^2 A + 1 = \sec^2 A \notag \]
Use a strategic division and simplification to derive \(1 + \cot^2 A = \csc^2 A\) from the original \( \sin^2 A + \cos^2 A = 1\).
- Answer
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Hint: what do you need to divide everyone by in order to make the first term into a \(1\)?
Note that \(A\) can be anything in these identities, as long as it's consistent in the terms. You can also rearrange the equation in any way you need, legally. For example,
\[ \sin^2 x + \cos^2 x = 1, \quad \quad \cos^2 (a+b) + \sin^2 (a+b) = 1, \quad \quad 1 - \sin^2 (2\theta) = \cos^2 (2\theta) \notag \]
are all true, but
\[ \sin^2 \theta - \cos^2 \theta \neq 1, \quad \quad \sin^2 x + \cos^2 3x \neq 1, \quad \quad \sin x + \cos x \neq 1 \notag \]
and so on. Make sure you look at each example there and understand why.
Even/Odd Identities
Since \( \sin x\) is symmetric about the origin, it's an odd function (review Section 4.2 if needed). Since \( \cos x\) is symmetric across the \(y\)-axis, it's an even function. These facts give us the following identities:
\[ \sin(-x) = - \sin x \quad \text{ and } \quad \cos(-x) = \cos x \notag \]
Find an even/odd identity for tangent by considering what \( \tan (-x) = ...\) turns out to be.
- Write tangent in terms of sines and cosines.
- Apply the two facts above and simplify the result.
- Answer
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\( \tan(-x) = - \tan x\).
Sum/Difference Identities
I actually saw these come up the last time I taught Calc I, so I want you to be familiar with the identities involving sine and cosine below, but I don't expect you to memorize the ones involving tangent (ask your professor if they want you to!). These allow you to rewrite trig functions with complicated inputs as expressions involving simpler inputs.
For sine, the formulas for addition or subtraction in the input have mixed terms and matching operation:
\[ \sin(A \textcolor{red}{+}B) = \sin A \cos B \textcolor{red}{+} \cos A \sin B, \quad \quad \sin(A \textcolor{red}{-}B) = \sin A \cos B \textcolor{red}{-} \cos A \sin B \notag \]
For cosine, we have matching terms and opposite operation:
\[ \cos( A \textcolor{red}{+} B) = \cos A \cos B \textcolor{red}{-} \sin A \sin B, \quad \quad \cos (A \textcolor{red}{-} B) = \cos A \cos B \textcolor{red}{+} \sin A \sin B \notag \]
Also, for tangent, if needed, we have:
\[ \tan(A + B) = \frac{ \tan A + \tan B}{1- \tan A \tan B}, \quad \quad \tan(A-B) = \frac{ \tan A - \tan B}{1 + \tan A \tan B} \notag \]
I've done some strategic color-coding above to help you notice important features while memorizing. Once again, if you find yourself forgetting the subtraction formulas above, you can actually just deduce them from the addition versions, using the even/odd identities!
Prove the difference formula for sine using the addition formula \(\sin(A+B) = \sin A \cos B + \cos A \sin B \) and the even/odd identities.
Solution
We want to figure out how to rewrite \( \sin(A - B) \), so we start with that and try manipulating it. The first thing to observe is that \( A - B = A + (-B) \). Now we can apply the sum formula given above, treating \(A\) and \( (-B)\) as the added components.
\[ \sin(A+\textcolor{red}{(-B)}) = \sin A \cos \textcolor{red}{(-B)} + \cos A \sin \textcolor{red}{(-B)} \notag \]
Next, we apply the facts that \( \cos (-x) = \cos x\) and \( \sin(-x) = -\sin x\), to get
\[ \sin(A - B) = \sin A \cos B + (\cos A )(- \sin B) = \sin A \cos B - \cos A \sin B \notag \]
Now, that was with a lot of explanation, but let's say I asked you to "prove" something like this on a test. It would be sufficient (in my opinion, I'm a chill grader) to just write down a true chain of equations showing all the intermediate steps, like this:
\[ \sin(A-B) = \sin(A + (-B)) = \sin A \cos(-B) + \cos A \sin(-B) = \sin A \cos B - \cos A \sin B \notag \]
But of course, since the idea is to show why something is true, you cannot leave out any of those intermediate steps!
Prove the difference formula for cosine using the addition formula \( \cos( A + B) = \cos A \cos B - \sin A \sin B \) and the even/odd identities.
- Answer
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\[ \cos (A - B) = \cos (A + (-B)) = \cos A \cos(-B) - \sin A \sin (-B) = \cos A \cos B + \sin A \sin B \notag \]
Double Angle Identities
The name double angle means that these formulas are used when the input for a trig function is "\( 2\) times something" specifically.
For sine, we have
\[ \sin 2A = 2 \sin A \cos A \notag \]
For cosine, we have several equivalent expressions
\[ \cos 2A = \textcolor{red}{\cos^2 A - \sin^2 A} = \textcolor{blue}{ 1 - 2 \sin^2 A} = \textcolor{green}{2 \cos^2 A - 1} \notag \]
Also, if needed, we have for tangent
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 x} \notag \]
Let's prove some of these double angle formulas using the sum/difference identities we already learned. (Are you sensing a theme here?)
Prove the double angle identities for \( \cos 2A\).
Solution
The key here is to recognize that \(2A = A + A\). Using the sum formula with \(A\) plugged in for both \(A\) and \(B\), we have
\[ \cos 2A = \cos(A+A) = \cos A \cos A - \sin A \sin A = (\cos A)^2 -(\sin A)^2 = \cos^2 A - \sin^2 A \notag \]
Great, but how do we show that the other expressions are equivalent to the one above? For this, we're going to use the Pythagorean Identity \( \sin^2 A + \cos^2 A = 1\). First, we rearrange it to solve for \( \cos^2 A\):
\[ \cos^2 A = 1 - \sin^2 A \notag \]
Then we go back to our double-angle identity expression and replace \( \cos^2 A\) with what he's equal to, and simplify:
\[\cos^2 A - \sin^2 A = (1 - \sin^2 A) - \sin^2 A = 1 - \sin^2 A - \sin^2 A = 1 - 2\sin^2 A \notag \]
To get the final one, we instead solve the Pythagorean Identity for the \(\sin^2 A \) term:
\[ \sin^2 A = 1 - \cos^2 A \notag \]
And then do the same substitution trick (use parentheses!):
\[ \cos^2 A - \sin^2 A = \cos^2 A - (1 - \cos^2 A) = \cos^2 A - 1 + \cos^2 A = 2\cos^2 A - 1 \notag \]
Since everybody is equal to \(\cos^2 A - \sin^2 A\) and that guy is equal to \( \cos 2A\), we're done!
Prove the double-angle formula for \( \sin 2A\) using the sum formula we learned above (this is less work than the example we just did).
- Answer
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\[ \sin 2A = \sin(A+A) = \sin A \cos A + \sin A \cos A = 2 \sin A \cos A \notag \]
I want to emphasize that you should know these identities forwards and backwards! Here's an example of using a double-angle formula backwards...
Evaluate \( \sin\left( \frac{\pi}{8}\right)\cos\left( \frac{\pi}{8}\right) \) even though \( \pi/8\) isn't a "typical" angle.
Solution
We notice that the expression is of the form \( \sin A \cos A \), which we know appears on the right hand side of the double angle identity
\[ \sin 2A = 2 \sin A \cos A. \notag \]
We can rearrange this identity to become
\[ \sin A \cos A = \frac{1}{2} \sin(2A). \notag \]
Now, letting \(A = \frac{\pi}{8}\), we have
\[ \sin\left( \frac{\pi}{8}\right)\cos\left( \frac{\pi}{8}\right) = \frac{1}{2} \sin\left( \frac{2\pi}{8}\right) = \frac{1}{2} \sin\left( \frac{\pi}{4}\right), \notag \]
which evaluates to \( \frac{1}{2\sqrt{2}} \).
These aren't the only trig identities that people study in something like a dedicated Analytical Trigonometry course, and you may see more crop up from time to time in specific contexts. These include things called cofunction identities, half-angle formulas, product-sum formulas, and whatever other fun little equivalences people prove and write down. The ones I've chosen to include in this class are the ones that I feel were most helpful to have on instant recall throughout my undergraduate math career. If you're interested in other identities, we'll explore some of them in the exercises, and there are a lot of nice "cheat sheet" lists out there, such as this one.
I hope by this point in the course you're starting to notice that studying math is not like other subjects where you have to memorize a ton of seemingly random factoids, like dates and dead people, or Latin names for things... Math builds on itself, and we use math facts to deduce or prove other math facts. Sure, you need to memorize a few facts to start out, but once you have a basic toolbox, you can derive everything you need using logic. The best thing you can take away from this class is the ability to puzzle out the consequences of facts, or know when something has been logically justified from start to finish. Like how we used the even/odd identities to figure out the difference formulas, or used the sum formulas to figure out the double-angle formulas. Math is not black magic and chaos that you have no hope of retaining. Really try to engage with the WHY and HOW behind anything you're learning; that synthesis and integration is what learning is all about!
Anyways, all this to say, go do the exercises to explore utilizing all these facts.