Trigonometric fundamental identities are really fun stuff. We can use them to rewrite equations and make them simpler for us to solve, and I will illustrate this further along in this post.
Let’s get into the Pythagorean identities first.
Pythagorean Identities
1) \ \text{sin}^2\theta + \text{cos}^2\theta=1 \\ 2) \ \text{tan}^2\theta + 1 = \text{sec}^2\theta \\ 3) \ \text{cot}^2\theta+1=\text{csc}^2\theta=1Now that we have our three Pythagorean identities written out, let’s also make note of how these identities can be rewritten.
1) \ \text{cos}^2\theta=1-\text{sin}^2\theta \\2) \ \text{sec}^2\theta-1=\text{tan}^2\theta \\3) \ \text{csc}^2\theta-\text{cot}^2\theta = 1These Pythagorean identities will be critical to your success, so make sure to memorize them.
Even Functions
What is an even function? An even function is one that remains symmetrical with respect to the y-axis, meaning that if you reflect the function across the y-axis, it will remain unchanged.
Here are our even functions.
\text{cos}(-\theta)=\text{cos}\theta \\ \text{sec}(-\theta) = \text{sec}\thetaOdd Functions
What is an odd function? Odd functions are functions that remain symmetrical with respect to the origin, meaning that if you rotate the graph of the function 180 degrees around the origin, the graph will remain unchanged.
Here are the odd functions.
\text{sin}(-\theta)=-\text{sin}\theta \\ \text{csc}(-\theta) = -\text{csc}\theta \\ \text{tan}(-\theta)=-\text{tan}\theta \\ \text{cot}(-\theta) = -\text{cot}\thetaTests
- If f(x) = f(-x), then it’s an even function.
- If f(-x) = -f(x), then it’s an odd function.
Examples
\text{If sin}\theta = \frac{3}{5}, \text{ then -sin}\theta = \\ -\text{sin}(-\theta) \\ = -(-\text{sin}\theta) \\ =-(-\frac{3}{5}) \\ =\frac{3}{5}How to algebraically determine if a function is even or odd
f(x) = \frac{\text{secx}}{\text{x}} \\ f(-x)=\text{sec(-x)} \\ f(-x) = \frac{\text{sec-x}}{\text{-x}} \\ f(-x) = \frac{-\text{secx}}{x} \\ \text{not even} \\ -f(x) = \frac{-secx}{x} \\ \text{so it's an odd function}In order to prove whether a function is even or odd, we must take the function and plug in -x for x. If you end up with the same function you started with, then the function is even, and if you end up with the opposite, then your function is odd.
At the top, we have the function that we are trying to determine if it is even, odd or neither.
f(x) = \frac{\text{secx}}{\text{x}}First, we must plug in -x for x.
f(-x) = \text{sec(-x)} \\ f(-x) = \text{sec-x}Then we are going to divide the function by -x.
f(-x) = \frac{\text{sec-x}}{\text{-x}}Secant is going to become negative, x is going to become positive in both the numerator and the denominator because two negatives make a positive.
f(-x) = \frac{\text{-secx}}{\text{x}}There is no further way to manipulate this equation, so we see that we ended with a different equation than we started with. This equation is not even, so it is odd.
Let’s take a look at our next example.
Find the 6 trig functions given the following information and using the pythagorean identities
\text{tan}\alpha = -\frac{1}{2}, \alpha \text{ is in quadrant 4} Right away, we can find our cotangent first, because that’s just the reciprocal of the tangent.
\text{cot}\alpha = -2Remember that in Quadrant 4, cosine and secant are positive, and the rest are negative. Click here to see an easy way to remember the signs in each quadrant.
Let’s use one of our Pythagorean identities to solve this.
\text{tan}^2\theta + 1=\text{sec}^2\theta Since we are given tan alpha, let’s substitute that into the identity.
(-\frac{1}{2})^2+1=\text{sec}^2\theta\frac{1}{4}+\frac{4}{4} = \text{sec}^2\theta\text{sec}^2\theta = \frac{5}{4}\text{sec}\theta =\pm\sqrt{\frac{5}{4}}\text{sec}\theta=\frac{\sqrt{5}}{2}Now to find cosine theta, we can simply take the reciprocal of the secant. We know that cosine is positive in quadrant 4, so our answer is going to be positive.
\text{cos}\theta=\frac{2}{\sqrt{5}} \cdot\frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}Now we can take this information and plug it into another Pythagorean identity to find sine theta.
\text{cos}^2\theta+\text{sin}^2\theta=1 \\ (\frac{2\sqrt{5}}{5})^2+\text{sin}^2\theta = 1 (When we square the first value, we are essentially saying 2 times 2 times square root 5 times square root 5 – or just 5, because a square root times itself is just that number without the square roots)
\frac{20}{25}+\text{sin}^2\theta=1Now subtract 20/25 from both sides of the equation
\text{sin}^2\theta=\frac{25}{25}-\frac{20}{25} \text{sin}^2\theta = \frac{5}{25} \\ \text{sin}^2\theta= \frac{1}{5}\text{sin}\theta = \pm\sqrt{\frac{1}{5}}We know that sine is negative in quadrant 4, so our answer is going to be negative.
\text{sin}\theta=\frac{1}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=-\frac{\sqrt{5}}{5}Now we can take the reciprocal of sine and find our cosecant.
\text{csc}=-\sqrt{5}So now that we have fully worked through our problem and gotten all our answers, let’s reiterate what all the 6 trig functions are.
\text{tan}\alpha = -\frac{1}{2} \ \ \ \ \ \text{cot}\alpha = -2 \\ \text{sec}\alpha=\frac{\sqrt{5}}{2} \ \ \ \ \ \text{cos}\alpha = \frac{2\sqrt{5}}{5} \\ \ \ \ \ \ \text{sin}\alpha = -\frac{\sqrt{5}}{5} \ \ \ \ \ \text{csc}\alpha=-\sqrt{5}Find the 6 trig functions given the following information
\text{sec}\theta = \frac{9}{8}, \text{sin}\theta < 0We need to figure out what quadrant we are working in. We know that secant is positive in quadrants 1 and 4, and sine is negative in quadrants 3 and 4, so the only quadrant that works to make both statements true is quadrant 4.
Now, I like to draw a picture of the problem because it makes this much easier to solve.
Now I knew where to put 8 and 9 because the reciprocal of secant is cosine, and according to SOH CAH TOA, cosine is adjacent over hypotenuse.
This is how I filled in the last part of the triangle with negative square root 17.
y=-\sqrt{(9)^2+(8)^2} \\ y=-\sqrt{81-64} \\ y = -\sqrt{17}Now, using SOH CAH TOA, we can just plug these numbers in for the rest of our answers!
\text{cos}\theta=\frac{8}{9} \ \ \ \ \ \text{sin}\theta=-\frac{\sqrt{17}}{9}\text{csc}\theta=-\frac{9}{\sqrt{17}} = -\frac{9\sqrt{17}}{17}\text{tan}\theta=-\frac{\sqrt{17}}{8}\text{cot}\theta = -\frac{8}{\sqrt{17}} = -\frac{8\sqrt{17}}{17}Write the expression in terms of sine and cosine only
\text{sec}\theta\text{sin}\theta\text{cot}\theta=\frac{1}{\cancel{cos\theta}} \cdot\frac{{\cancel{sin\theta}}}{1}\cdot\frac{\cancel{cos\theta}}{\cancel{sin\theta}} = 1FOIL and simplify this expression
(\text{csc}\theta+\text{sec}\theta)(\text{sin}\theta-\text{cos}\theta)=\text{csc}\theta\text{sin}\theta-\text{cos}\theta\text{csc}\theta+\text{sec}\theta\text{sin}\theta-\text{cos}\theta\text{sec}\theta =\frac{1}{\cancel{sin\theta}} \cdot \frac{\cancel{sin\theta}}{1} - \frac{cos\theta}{1} \cdot \frac{1}{sin\theta} + \frac{1}{cos\theta} \cdot \frac{sin\theta}{1} - \frac{cos\theta}{1} \cdot \frac{1}{cos\theta}=1 - \frac{cos\theta}{sin\theta} + \frac{sin\theta}{cos\theta} - 1= -cot\theta + tan\theta \\ =tan\theta-cot\theta
Simplify this expression
\text{cos}\theta(\text{sec}\theta-\text{cos}\theta)\frac{\cancel{cos\theta}}{1} \cdot \frac{1}{\cancel{cos\theta}} - cos^2\theta \\ 1-cos^2\theta \\ =sin^2\thetaSimplify this expression
\frac{1+cot^2\theta}{csc^2\theta-1}*Remember –
cot^2\theta + 1 = csc^2\theta
We are going to use this identity above to simplify our expression.
=\frac{csc^2\theta}{cot^2\theta} \\=\frac{1}{sin^2\theta} \div \frac{cos^2\theta}{sin^2\theta} \\ =\frac{1}{sin^2\theta}\cdot\frac{sin^2\theta}{cos^2\theta} \\ =\frac{1}{cos^2\theta} \\=sec^2\thetaConclusion
I know this was a very lengthy post, and if you made it through all this, congratulations. I hope you found this post to be helpful and helped you to make sense of trigonometric identities. If you did find this helpful (or if you didn’t) please leave me a comment!