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SectionIntroduction to Trigonometric Identities

In this section, we will introduce the concept of trigonometric identities and work with several identities to prove that they are true.

SubsectionTrigonometric Identities

A trigonometric identity is a trigonometric equation that is true for every possible value of the input variable on which it is defined.

Identities are usually something that can be derived from definitions and relationships we already know. One identity that we are already familiar with is the Pythagorean Identity, which we derived from the definitions of sine and cosine. Recall the Pythagorean Identity states that for any angle \(\theta\text{,}\)

\begin{equation*} \cos^2(\theta)+\sin^2(\theta)=1 \end{equation*}

In some cases, we can use trigonometric identities to simplify an expression. To do so, we can utilize the definitions and identities we have already established.


Simplify the expression \(\displaystyle \frac{\sec(\theta)}{\tan(\theta)}\text{.}\)


We begin by writing secant and tangent in terms of sine and cosine. This yields

\begin{equation*} \frac{\sec(\theta)}{\tan(\theta)} = \frac{\frac{1}{\cos(\theta)}}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{1}{\cos(\theta)}\cdot\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)} = \csc(\theta). \end{equation*}

Sometimes a question may ask you to "prove the identity" or "establish the identity." This is the same idea as when you are asked to show that \((x-1)^2 = x^2-2x+1\text{.}\) In this type of question, we must show the algebraic manipulations that demonstrate that the left and right side of the equation are in fact equal. You can think of a "prove the identity" problem as a simplification problem where you know the answer: you know what the end goal of the simplification should be and you just need to show the steps to get there.

In most cases, to prove an identity you will start with the expression on one side of the identity and manipulate it using algebra and trigonometric identities until you have simplified it to the expression on the other side of the equation. Note that we do not treat the identity like an equation to solve - it isn't! Instead, we are trying to prove if the two expressions are equal, so we must take care to work with one side at a time rather than applying an operation simultaneously to both sides of the equation.


Prove the identity \(\displaystyle \frac{1+\cot(\theta)}{\csc(\theta)} = \sin(\theta) + \cos(\theta)\text{.}\)


We start with the left-hand side of the desired identity and try to manipulate it to look like the right-hand side. By writing cotangent and cosecant in terms of sine and cosine, we have

\begin{align*} \frac{1+\cot(\theta)}{\csc(\theta)} \amp = \frac{1+\frac{\cos(\theta)}{\sin(\theta)}}{\frac{1}{\sin(\theta)}}\\ \\ \amp = \left( 1+\frac{\cos(\theta)}{\sin(\theta)} \right) \cdot \sin(\theta)\\ \\ \amp = \sin(\theta)+\frac{\cos(\theta)}{\sin(\theta)}\cdot\sin(\theta)\\ \\ \amp = \sin(\theta)+\cos(\theta). \end{align*}

We can also use identities that we have previously learned, like the Pythagorean Identity, while simplifying or proving identities.


Establish the identity \(\displaystyle \frac{\cos^2(\theta)}{1+\sin(\theta)} = 1-\sin(\theta)\text{.}\)


To establish this identity, let's manipulate both sides of the given identity until we find an identity we know to be true. Begin with

\begin{equation*} \frac{\cos^2(\theta)}{1+\sin(\theta)} = 1-\sin(\theta). \end{equation*}

Let's multiply both sides by \(1+\sin(\theta)\text{,}\) which gives

\begin{equation*} \cos^2(\theta) = (1-\sin(\theta))\cdot(1+\sin(\theta)). \end{equation*}

Next, we multiply out the right-hand side, which gives

\begin{equation*} \cos^2(\theta) = 1+\sin(\theta)-\sin(\theta)-\sin^2(\theta). \end{equation*}

Canceling like terms on the right-hand side, we have

\begin{equation*} \cos^2(\theta)=1-\sin^2(\theta). \end{equation*}

We know the last line to be true due to the Pythagorean identity, \(\cos^2(\theta)+\sin^2(\theta)=1\text{.}\) Therefore, the initial identity must also be true.

We can also build new identities from previously established identities. For example, we can come up with another identity if we divide both sides of the Pythagorean Identity by cosine squared (which is allowed since we've already shown the identity is true).

\begin{align*} \frac{\cos^2(\theta)+\sin^2(\theta)}{\cos^2(\theta)} \amp = \frac{1}{\cos^2(\theta)} \amp\amp \text{Splitting the fraction on the left} \\ \\ \frac{\cos^2(\theta)}{\cos^2(\theta)} + \frac{\sin^2(\theta)}{\cos^2(\theta)} \amp = \frac{1}{\cos^2(\theta)} \amp\amp \text{Simplifying} \\ \\ 1 + \frac{\sin^2(\theta)}{\cos^2(\theta)} \amp = \frac{1}{\cos^2(\theta)} \amp\amp \text{Using the definition } \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \\ \\ 1 + \tan^2(\theta) \amp = \frac{1}{\cos^2(\theta)} \amp\amp \text{Using the definition } \sec(\theta) = \frac{1}{\cos(\theta)} \\ \\ 1 + \tan^2(\theta) \amp = \sec^2(\theta) \amp\amp \end{align*}

Use a similar approach to establish that \(\cot^2(\theta)+1=\csc^2(\theta)\text{.}\)

Alternate Forms of the Pythagorean Identity

\begin{align*} 1 + \tan^2(\theta) \amp = \sec^2(\theta)\\ \\ \cot^2(\theta) + 1 \amp = \csc^2(\theta) \end{align*}

SubsectionSolving Trigonometric Equations with Identities

In the last chapter, we solved basic trigonometric equations. Using identities, we can now explore different techniques to solve more complicated trig equations. Building from what we already know makes this a much easier task.

Consider the function \(f(x) = 2x^2+x\text{.}\) If you were asked to solve \(f(x) = 0\text{,}\) you could use algebra to arrive at two solutions.

First, we set \(\ 2x^2+x = 0 \ \) and factor, which gives us

\begin{equation*} x(2x + 1) = 0 \end{equation*}

Now, since the terms \(\ x \ \) and \(\ 2x+1 \ \) multiply together to equal 0, we know that either \(\ x=0 \ \) or \(\ 2x+1=0 \ \text{.}\) Therefore, we get the two solutions

\begin{equation*} x=0 \hspace{.25in} \text{ and } \hspace{.25in} x=-\frac{1}{2} \end{equation*}

Following the same steps, we can solve the equation \(f(\theta) = 0\) where \(f(\theta) = 2\sin^2(\theta) + \sin(\theta)\text{.}\) We get that

\begin{align*} 2\sin^2(\theta) + \sin(\theta) \amp = 0 \amp\amp \text{Factoring out } \sin(\theta) \\ \\ \sin(\theta)\Big(2\sin(\theta) + 1\Big) \amp = 0 \amp\amp \end{align*}

Like before, we have two terms that multiply together to get 0. Therefore, we know that either

\begin{equation*} \sin(\theta) = 0 \hspace{.25in} \text{ or } \hspace{.25in} 2\sin(\theta) + 1 = 0 \end{equation*}

We must now solve each of these equations to get a complete set of solutions.

Solving the equation \(\ \sin(\theta) = 0 \ \) gives us the solutions \(...,-\pi,0,\pi,2\pi,...\) which can be represented as the solution set

\begin{equation*} \theta = \pi k \hspace{.25in} \text{where } k \text{ is any integer} \end{equation*}

Solving the equation \(\ 2\sin(\theta) + 1 = 0 \) we get \(\displaystyle \sin(\theta) = -\frac{1}{2} \text{,}\) which gives us the solution set

\begin{equation*} \theta = \frac{7\pi}{6}+2\pi k \hspace{.25in} \text{and} \hspace{.25in} \theta = \frac{11\pi}{6} + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \end{equation*}

Therefore, the solutions to the original equation \(\ 2\sin^2(\theta) + \sin(\theta) = 0 \ \) are

\begin{align*} \theta \amp = \pi, \amp\amp \\ \\ \theta \amp = \frac{7\pi}{6}+2\pi k, \amp\amp \text{and} \\ \\ \theta \amp = \frac{11\pi}{6} + 2\pi k \amp\amp \text{ where } k \text{ is any integer} \end{align*}

Solve \(\ 3\sec^2(\theta) - 5\sec(\theta) = 2 \ \) for all solutions with \(0 \leq \theta \lt 2\pi\text{.}\)


Consider the function \(f(x)=3x^2-5x\text{.}\) If you were asked to find all solutions to \(f(x)=2\text{,}\) you could use algebra to arrive at two solutions:

\begin{align*} 3x^2-5x \amp = 2 \\ 3x^2-5x-2 \amp = 0 \\ (x-2)(3x+1) \amp = 0. \end{align*}

Therefore, we get the two solutions

\begin{equation*} x=2 \text{ and } x=-\frac{1}{3}. \end{equation*}

Following these steps as above with \(\sin(\theta)\) instead of \(x\text{,}\) we know that either

\begin{equation*} \sin(\theta)=2 \text{ or } \sin(\theta)=-\frac{1}{3}. \end{equation*}

There are no solutions to the equation \(\sin(\theta)=2\text{.}\) Solving the equation \(\sin(\theta)=-\frac{1}{3}\text{,}\) we get two solution sets. First,

\begin{align*} \theta \amp = \arcsin\left(-\frac{1}{3}\right) + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer} \\ \amp = -0.340 + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer}. \end{align*}

The other solution set for \(\sin(\theta)=-\frac{1}{3}\) has base angle \(\pi-\arcsin\left(-\frac{1}{3}\right)\text{,}\) which gives

\begin{equation*} \theta = 3.481 + 2\pi k \hspace{.25in} \text{ where } k \text{ is any integer}. \end{equation*}

Finally, we need to determine which solutions lie in the interval \(0\le\theta\le 2\pi\text{.}\) This happens for \(k=1\) in the first solution set and for \(k=0\) in the second solution set. Therefore, solutions to the original equation \(3\sec^2(\theta) - 5\sec(\theta) = 2\) in the interval \([0,2\pi]\) are

\begin{align*} \theta \amp = -0.340 + \pi = 2.801\\ \\ \theta \amp = 3.481. \end{align*}

Solve \(\ 2\sin^2(\theta)+3\sin(\theta) = -1 \ \) for all solutions with \(0 \leq \theta \lt 2\pi\text{.}\)

SubsectionThe Double Angle Identities

The Double Angle Identities

\begin{align*} \sin(2\theta) \amp = 2\sin(\theta)\cos(\theta) \\ \\ \cos(2\theta) \amp = \cos^2(\theta) - \sin^2(\theta) \end{align*}