###### Supplemental Videos

The main topics of this section are also presented in the following videos:

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The main topics of this section are also presented in the following videos:

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

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

Solution

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.

There are two primary ways to prove identities:

Start with a known identity and algebraically manipulate both sides to get a new identity.

Start with one side of an identity you want to prove, then manipulate it using known identities to get the other side.

In this section, we will show examples using both strategies. It's important to note that there is no "standard algorithm" for proving identities! Every problem is a little different, but as you prove more identities you may begin to notice patterns that you can often use (such as writing tangent in terms of sine and cosine).

Something else to keep in mind is that when we are asked to prove a trigonometric identity, we do no treat the identity like an equation to solve - it isn't! Instead, we are trying to prove that 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 \(\displaystyle \frac{1+\cot(\theta)}{\csc(\theta)} = \sin(\theta) + \cos(\theta)\text{.}\)

Solution

To prove this identity, we will use the second strategy: we will start with the left-hand side 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*}

Since we have shown that the left-hand side is equal to the desired right-hand side, we have proven the identity.

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

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

Solution

To establish this identity, we will use the first strategy: we will start with a known identity and manipulate both sides to get the identity we want. In particular, let's start with the Pythagorean Identity:

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

In proving identities, it's important to keep in mind "where we're going": we're trying to show that \(\frac{\cos^2(\theta)}{1+\sin(\theta)}\) is equal to something, so it would be helpful at this point to divide both sides by \(1+\sin(\theta)\text{:}\)

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

We have the left-hand side of the identity we want; we now just need to manipulate the right-hand side to be \(1-\sin(\theta)\text{.}\) We can factor \(1-\sin^2(\theta)\) in order to cancel the denominator:

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

Since we have shown that the left-hand side of the identity is equal to the right-hand side, we have proven the identity.

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*}

Therefore \(1+\tan^2(\theta)=\sec^2(\theta)\) is another trigonometric identity.

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

Solution

We begin by dividing both sides of the Pythagorean identity by \(\sin^2(\theta)\text{,}\) then follow the same steps as from above:

\begin{align*}
\frac{\cos^2(\theta)+\sin^2(\theta)}{\sin^2(\theta)} \amp = \frac{1}{\sin^2(\theta)}\\
\\
\frac{\cos^2(\theta)}{\sin^2(\theta)} + \frac{\sin^2(\theta)}{\sin^2(\theta)} \amp = \frac{1}{\sin^2(\theta)}\\
\\
\frac{\cos^2(\theta)}{\sin^2(\theta)} + 1 \amp = \frac{1}{\sin^2(\theta)}\\
\\
\cot^2(\theta) + 1 \amp = \frac{1}{\sin^2(\theta)}\\
\\
\cot^2(\theta) + 1 \amp = \csc^2(\theta).
\end{align*}

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

Another useful type of trigonometric identities are the double angle identities. The double angle identities allow us to take trigonometric equations with \(2\theta\) and replace them with trigonometric expressions containing just \(\theta\text{.}\) We will first derive the double angle formula for \(\sin(2\theta)\text{,}\) and then the double angle formula for \(\cos(2\theta)\text{.}\)

The first double angle identity that we will prove is

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

which we will prove using the Law of Sines. We start by drawing a right triangle with one of the angles equal to \(\theta\) and a hypotenuse of length \(1\text{.}\) We will label the other sides \(x\) and \(y\) and the other angle \(\alpha\) as shown below.

Because this is a right triangle, we know that

\begin{equation*}
\begin{aligned}
\cos(\alpha)=\dfrac{y}{1}=y\hspace{.5cm}\amp\text{and}\hspace{.5cm}\sin(\theta)=\dfrac{y}{1}=y \\
\sin(\alpha)=\dfrac{x}{1}=x\hspace{.5cm}\amp\text{and}\hspace{.5cm}\cos(\theta)=\dfrac{x}{1}=x
\end{aligned}
\end{equation*}

Next, we attach a copy of this triangle to itself as though it has been reflected across the side labeled \(x\text{:}\)

Together, these two triangles form one large isosceles triangle; by the Law of Sines, this triangle gives

\begin{equation*}
\dfrac{\sin(\alpha)}{1}=\dfrac{\sin(2\theta)}{2y}
\end{equation*}

or

\begin{equation*}
\sin(2\theta)\cdot1=2y\cdot\sin(\alpha)
\end{equation*}

Using the fact that \(\sin(\alpha)=x=\cos(\theta)\) and \(y=\sin(\theta)\text{,}\) this becomes simply

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

as desired.

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

The second double angle identity is

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

and can be proved similarly, using the Law of Cosines instead of the Law of Sines. We will reuse the same triangles as above:

By the Law of Cosines,

\begin{equation*}
\begin{aligned}
(2y)^2 \amp= 1^2+1^2-2(1)(1)\cos(2\theta) \\
4y^2 \amp= 2-2\cos(2\theta) \\
\cos(2\theta) \amp= 1-2y^2
\end{aligned}
\end{equation*}

But from the original right triangles, we know that \(y=\sin(\theta)\text{,}\) so \(y^2=\sin^2(\theta)\text{.}\) This, combined with the identity \(\cos^2(\theta)=1-\sin^2(\theta)\text{,}\) gives

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

as claimed.

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

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

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.

Solve the trigonometric equation \(\cos^2(\theta)-2\sin(\theta)+2=0\) for all values of \(\theta\text{.}\)

Solution

Because this equation has both the square of cosine and a sine, we cannot factor this like we have in the past. However, we have a trigonometric identity for \(\cos^2(\theta)\) in terms of sine:

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

Substituting this into our equation, we can simplify and then factor.

\begin{equation*}
\begin{aligned}
\cos^2(\theta)-2\sin(\theta)+2 \amp= 0 \\
\alert{1-\sin^2(\theta)}-2\sin(\theta)+2 \amp= 0 \\
-\sin^2(\theta)-2\sin(\theta)+3 \amp= 0 \\
(-\sin(\theta)+1)(\sin(\theta)+3) \amp= 0
\end{aligned}
\end{equation*}

This means that either \(-\sin(\theta)+1=0\) or \(\sin(\theta)+3=0\text{.}\) It cannot be the case that \(\sin(\theta)+3=0\) since there is no solution to \(\sin(\theta)=-3\text{.}\) The other equation, \(-\sin(\theta)+1=0\text{,}\) implies that \(\sin(\theta)=1\text{,}\) which gives an initial solution of \(\theta=\frac{\pi}{2}\text{.}\) Since sine is \(2\pi\)-periodic, the solution set is

\begin{equation*}
\theta=\frac{\pi}{2}+2\pi k \hspace{.5cm}\text{for any integer }k
\end{equation*}

Solve the trigonometric equation \(\sin(2\theta)+\cos(\theta)=0\) for all values of \(\theta\text{.}\)

Solution

We will use the double angle identity \(\sin(2\theta)=2\sin(\theta)\cos(\theta)\text{:}\)

\begin{equation*}
\begin{aligned}
\sin(2\theta)+\cos(\theta) \amp= 0 \\
\alert{2\sin(\theta)\cos(\theta)}+\cos(\theta) \amp= 0
\end{aligned}
\end{equation*}

We can now factor out \(\cos(\theta)\text{.}\)

\begin{equation*}
\begin{aligned}
2\sin(\theta)\cos(\theta)+\cos(\theta) \amp= 0 \\
\cos(\theta)\big(2\sin(\theta)+1\big) \amp= 0
\end{aligned}
\end{equation*}

This means that either \(\cos(\theta)=0\) or \(2\sin(\theta)+1=0\text{.}\) From the unit circle, if \(\cos(\theta)=0\) then we get two initial solutions:

\begin{equation*}
\theta=\dfrac{\pi}{2}\hspace{.5cm}\text{and}\hspace{.5cm}\theta=\dfrac{3\pi}{2}
\end{equation*}

The second case, \(2\sin(\theta)=1=0\text{,}\) implies that \(\sin(\theta)=-\frac{1}{2}\text{.}\) Again from the unit circle, we get two more initial solutions:

\begin{equation*}
\theta=\dfrac{7\pi}{6}\hspace{.5cm}\text{and}\hspace{.5cm}\theta=\dfrac{11\pi}{6}
\end{equation*}

Since both cosine and sine are \(2\pi\)-periodic, the solution set is

\begin{equation*}
\begin{aligned}
\theta \amp= \frac{\pi}{2}+2\pi k \\
\theta \amp= \frac{3\pi}{2}+2\pi k \\
\theta \amp= \frac{7\pi}{6}+2\pi k \\
\theta \amp= \frac{11\pi}{6}+2\pi k \hspace{.5cm}\text{for any integer }k
\end{aligned}
\end{equation*}