Section 4.2 Introduction to Trigonometric Identities

Supplemental Videos.

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.

Subsection 4.2.1 Trigonometric 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.

Example 4.7.

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.

Example 4.8.

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.

Example 4.9.

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.

Example 4.10.

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*}

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*}

Subsection 4.2.2 The Double Angle Identities

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*}

\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*}

Together, these form the 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*}

Subsection 4.2.3 Solving 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.

Example 4.11.

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*}

Example 4.12.

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*}

Exercises 4.2.4 Exercises

1. Manipulating Identities.

(a) Starting with the identity \(\sin^2(x)+\cos^2(x)=1\text{,}\) divide each side by \(\cos^2(x)\text{.}\) What is the resulting identity?

\(\displaystyle \tan^2(x)+1=\sec^2(x)\)
\(\displaystyle \sec^2(x)+\tan^2(x)=1\)
\(\displaystyle \cot^2(x)+1=\csc^2(x)\)
\(\displaystyle \csc^2(x)+1=\tan^2(x)\)
\(\displaystyle \tan^2(x)+\csc^2(x)=1\)
\(\displaystyle \tan^2(x)+1=\csc^2(x)\)
\(\displaystyle \csc^2(x)+\cot^2(x)=1\)
\(\displaystyle \sec^2(x)+1=\tan^2(x)\)
\(\displaystyle \csc^2(x)+1=\cot^2(x)\)
None of the above

(b) Let’s prove the identity \(\sin^2(x)+(\sec(x)-\cos(x))^2=\tan^2(x)\text{.}\) What is a good first step to show both sides are equal?

\(\displaystyle \sin^2(x)+\sec^2(x)+2\sec(x)\cos(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle \sin^2(x)+\sec^2(x)-2\sec(x)\cos(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle \sin^2(x)+\sec^2(x)+\cos^2(x)=\tan^2(x)\)
None of the above

\(\displaystyle -\cos^2(x)+\sec^2(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle \sec^2(x)+2\sec(x)\cos(x)=\tan^2(x)\)
\(\displaystyle \sin^2(x)+\sec^2(x)-2\sec(x)\cos(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle 1+\sec^2(x)+2\sec(x)\cos(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle 1+\sec^2(x)-2\sec(x)\cos(x)=\tan^2(x)\)
None of the above

(d) Using the definition of \(\sec(x)\text{,}\) we can rewrite the previous equation as...

\(\displaystyle 1+\sec^2(x)+2\tan(x)=\tan^2(x)\)
\(\displaystyle \sec^2(x)+2\sec(x)\cos(x)=\tan^2(x)\)
\(\displaystyle -\cos^2(x)+\tan^2(x)+\cos^2(x)=\tan^2(x)\)
\(\displaystyle \sec^2(x)+2\cos^2(x)=\tan^2(x)\)
\(\displaystyle 1+\sec^2(x)+2=\tan^2(x)\)
\(\displaystyle 1+\sec^2(x)-2=\tan^2(x)\)
None of the above

(e) The previous simplifies to...

\(\displaystyle 1+\sec^2(x)=-\tan^2(x)\)
\(\displaystyle \sec^2(x)-1=\tan^2(x)\)
\(\displaystyle \sec^2(x)+2=\tan^2(x)\)
\(\displaystyle \tan^2(x)=\tan^2(x)\)
\(\displaystyle 3+\sec^2(x)=\tan^2(x)\)
\(\displaystyle \sec^2(x)=\tan^2(x)\)
None of the above

(f) How can we finish the proof?

This equation is only one off from the Pythagorean Identity, so we can apply that here.
Add one to both sides to obtain the identity we proved in part (a).
The equation is true because both sides are identical.
This equation is true by the definition of the tangent and secant functions.
This equation is the negative of the identity we proved in part (a)
None of the above

2. Manipulating Identities.

(a) Starting with the identity \(\sin^2(x)+\cos^2(x)=1\text{,}\) divide each side by \(\sin^2(x)\text{.}\) What is the resulting identity?

\(\displaystyle \sec^2(x)+\tan^2(x)=1\)
\(\displaystyle 1+\tan^2(x)=\sec^2(x)\)
\(\displaystyle 1+\cot^2(x)=\sec^2(x)\)
\(\displaystyle 1+\sec^2(x)=\tan^2(x)\)
\(\displaystyle 1+\sec^2(x)=\cot^2(x)\)
\(\displaystyle 1+\cot^2(x)=\csc^2(x)\)
\(\displaystyle \csc^2(x)+\cot^2(x)=1\)
\(\displaystyle 1+\csc^2(x)=\cot^2(x)\)
\(\displaystyle \cot^2(x)+\sec^2(x)=1\)
None of the above

(b)

Let’s prove the identity \(\cos^2(x)+(\csc(x)-\sin(x))^2=\cot^2(x)\text{.}\) What is a good first step to show both sides are equal?

\(\displaystyle \cos^2(x)+\csc^2(x)+\sin^2(x)=\cot^2(x)\)
\(\displaystyle \cos^2(x)+\csc^2(x)+2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)\)
\(\displaystyle \cos^2(x)+\csc^2(x)-2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)\)
None of the above

\(\displaystyle -\sin^2(x)+\csc^2(x)+\sin^2(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)-2\csc(x)\sin(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)+2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)\)
\(\displaystyle \csc^2(x)+2\csc(x)\sin(x)=\cot^2(x)\)
\(\displaystyle \cos^2(x)+\csc^2(x)-2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)\)
None of the above

(d) Using the definition of \(\csc(x)\text{,}\) we can rewrite the previous equation as...

\(\displaystyle -\sin^2(x)+\cot^2(x)+\sin^2(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)+2\cot(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)+2=\cot^2(x)\)
\(\displaystyle \csc^2(x)+2\sin^2(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)-2=\cot^2(x)\)
\(\displaystyle \csc^2(x)+2\csc(x)\sin(x)=\cot^2(x)\)
None of the above

(e) The previous simplifies to...

\(\displaystyle \csc^2(x)=\cot^2(x)\)
\(\displaystyle \csc^2(x)-1=\cot^2(x)\)
\(\displaystyle \cot^2(x)=\cot^2(x)\)
\(\displaystyle 3+\csc^2(x)=\cot^2(x)\)
\(\displaystyle 1+\csc^2(x)=-\cot^2(x)\)
\(\displaystyle \csc^2(x)+2=\cot^2(x)\)
None of the above

(f) How can we finish the proof?

This equation is true by the definition of the cotangent and cosecant functions.
This equation is the negative of the identity we proved in part (a)
The equation is true because both sides are identical.
Add one to both sides to obtain the identity we proved in part (a).
This equation is only one off from the Pythagorean Identity, so we can apply that here.
None of the above

3. Solving Equations with Double Angles.

Using a double angle formula, determine all solutions for \(3\sin(2\theta)-3\cos(\theta) =0\)

Solutions with base angle in quadrant I: \(\theta =\) + \(n\)

Solutions with base angle in quadrant II: \(\theta =\) + \(n\)

Solutions with base angle not in any quadrant: \(\theta =\) + \(n\)

where \(n\) is any integer, and all base angles are between \(0\) and \(2\pi\text{.}\)

Use radian measure and enter your answers using at least four significant digits.

4. Verifying Identities.

Let’s verify the identity \(\cos(2\theta)=1-2\sin^2(\theta)\text{.}\) What is a good first step?

\(\displaystyle 2\cos(\theta)\sin(\theta)= 1-2\sin^2(\theta)\)
\(\displaystyle \cos^2(\theta)-\sin^2(\theta)= 1-2\sin^2(\theta)\)
\(\displaystyle \cos^2(\theta)+\sin^2(\theta)= 1-2\sin^2(\theta)\)
\(\displaystyle 2\cos(\theta)= 1-2\cos^2(\theta)\)
\(\displaystyle 2\cos(\theta)= 1+2\cos^2(\theta)\)
\(\displaystyle \cos^2(\theta)\sin^2(\theta)= 1-2\sin^2(\theta)\)
None of the above

What can we do to simplify our last equation?

\(\displaystyle \cos^2(\theta)= \csc^2(\theta)-2\)
\(\displaystyle 2\cos(\theta)= 1-2\sin(\theta)\)
\(\displaystyle -2\cos^2(\theta)+2\cos(\theta)+1=0\)
\(\displaystyle \cos^2(\theta)+\sin^2(\theta)= 1\)
\(\displaystyle 2\cos^2(\theta)-2\cos(\theta)+1=0\)
\(\displaystyle \cos^2(\theta)= 1-3\sin^2(\theta)\)
None of the above

How can we finish the proof?

We can factor the polynomial to solve for all values of \(\theta\text{.}\)
Our original equation must also be true since it is equivalent to the Pythagorean Identity.
We know our last equation is true by the definition of cosecant.
Divide by \(\cos^2(\theta)\) to obtain an equation we know is true.
This equation is the square root of the Pythagorean Identity.
None of the above

5. Verifying Identities.

Let’s verify the identity \(\cos(2\theta)=2\cos^2(\theta)-1\text{.}\) What is a good first step?

\(\displaystyle 2\cos(\theta)= 2\sin^2(\theta)-1\)
\(\displaystyle \cos^2(\theta)+\sin^2(\theta)= 2\cos^2(\theta)-1\)
\(\displaystyle 2\cos(\theta)\sin(\theta)= 2\cos^2(\theta)-1\)
\(\displaystyle 2\cos(\theta)= 2\sin^2(\theta)+1\)
\(\displaystyle \cos^2(\theta)-\sin^2(\theta)= 2\cos^2(\theta)-1\)
\(\displaystyle \cos^2(\theta)\sin^2(\theta)= 2\cos^2(\theta)-1\)
None of the above

What can we do to simplify our last equation?

\(\displaystyle -\cos^2(\theta)-\sin^2(\theta)= -1\)
\(\displaystyle 2\sin^2(\theta)-2\sin(\theta)+1\)
\(\displaystyle \sin^2(\theta)= 2-\sec^2(\theta)\)
\(\displaystyle 2\sin^2(\theta)-2\sin(\theta)-1\)
\(\displaystyle 2\sin(\theta)= 2\cos(\theta)-1\)
\(\displaystyle \sin^2(\theta)= 3\cos^2(\theta)-1\)
None of the above

How can we finish the proof?

We know our last equation is true by the definition of secant.
We can factor the polynomial to solve for all values of \(\theta\text{.}\)
This equation is the square root of the Pythagorean Identity.
Divide by \(\sin^2(\theta)\) to obtain an equation we know is true.
We may multiply both sides by \(-1\) to get the Pythagorean Identity.
None of the above

6. Verifying Identities.

Let’s verify the identity \(\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}\text{.}\) What do we get if we rewrite the right-hand side using sine and cosine?

\(\displaystyle \tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{1-\sin^2(\theta)\cos^2(\theta)}\)
\(\displaystyle \tan(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{1-\frac{\sin(\theta)}{\cos(\theta)}}\)
\(\displaystyle \tan(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{1-\frac{\sin^2(\theta)}{\cos^2(\theta)}}\)
\(\displaystyle \tan(2\theta)=\frac{2\frac{\cos(\theta)}{\sin(\theta)}}{1-\frac{\cos^2(\theta)}{\sin^2(\theta)}}\)
None of the above

What happens when we multiply the right hand side by \(\cos^2(\theta)/\cos^2(\theta)\text{?}\)

\(\displaystyle \tan(2\theta)=\frac{\frac{2\sin(\theta)}{\cos(\theta)}}{\cos^2(\theta)-\sin^2(\theta)}\)
\(\displaystyle \tan(2\theta)=\frac{2\cos(\theta)}{\cos^2(\theta)-\cos^2(\theta)}\)
\(\displaystyle \tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{1-\frac{\sin(\theta)}{cos(\theta)}}\)
\(\displaystyle \tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)}\)
None of the above

How can we use double angle formulas to simplify the right hand side?

\(\displaystyle \tan(2\theta)=\frac{\sin(2\theta)}{\cos(2\theta)}\)
\(\displaystyle \tan(2\theta)=\frac{\tan(2\theta)}{\sin^2(\theta)+\cos(2\theta)}\)
\(\displaystyle \tan(2\theta)=\frac{\cos(2\theta)}{\sin(2\theta)}\)
\(\displaystyle \tan(2\theta)=\sin(2\theta)-\cos(2\theta)\)
None of the above

How can we finish the proof?

The previous equation is true by the definition of tangent.
The previous equation is true after multiplying both sides by tangent.
The previous equation is true by the definition of sine and cosine.
The previous equation is equivalent to the Pythagorean Theorem.
The previous equation is true by the definition of the double angle formulas.
None of the above

7. Solving Equations Using Identities.

Determine all solutions for the following equation. [Hint: use an identity to write this equation purely in terms of sine.] \(\cos^2(\theta)+9\sin(\theta) - 9 =0\)

Solutions have the form: \(\theta =\) + \(n\)

where \(n\) is any integer, and the base angle is between \(0\) and \(2\pi\text{.}\)

Use radian measure and enter your answers using at least four significant digits.

8. Solving Equations Using Identities.

Determine all solutions for the following equation. [Hint: use an identity to write this equation purely in terms of tangent.] \(\sec^2(\theta)+\tan^2(\theta) =11\)

Solutions in quadrant I have the form: \(\theta =\) + \(n\)

Solutions in quadrant IV have the form: \(\theta =\) + \(n\)

where \(n\) is any integer, and all base angles are between \(-\pi\) and \(\pi\text{.}\)

Use radian measure and enter your answers using at least four significant digits.

9. Solving Equations with Double Angles.

Determine all solutions for the following equation. \(1+\tan(2\theta) =0\)

Solutions have the form: \(\theta =\) + \(n\)

where \(n\) is any integer, and the base angles is between \(-\frac{\pi}{4}\) and \(\frac{\pi}{4}\text{.}\)

Use radian measure and enter your answers using at least four significant digits.

Note: A problem like this is usually made easier by using a double-angle formula. However, this is an exception to the rule - try solving this problem using just standard techniques for solving inverse trig equations, no trig identities necessary.

10. Solving Equations with Double Angles.

Suppose that, during an experiment, the temperature of a substance in degrees Celsius at time t may be modeled by the equation \(C(t)=\sin(2t)-\cos(t)\text{.}\) Find all values of t for which the temperature of the substance is equal to \(0^\circ\) Celsius. [Hint: Start by drawing a sketch of C(t).]

Solutions with base angle in quadrant I: \(t=\) + \(n\)

Solutions with base angle in quadrant II: \(t=\) + \(n\)

Solutions with base angle on an axis: \(t=\) + \(n\)

where \(n\) is any integer, and all base angles are between \(0\) and \(2\pi\text{.}\)

Use radian measure and enter your answers using at least four significant digits.

11. Verifying Identities.

Let’s verify the identity \(\sin(2\theta)=\frac{2\tan(\theta)}{\sec^2(\theta)}\text{.}\) What happens when we rewrite the right-hand side using only sine and cosine?

\(\displaystyle \sin(2\theta)=2\frac{\cos^2(\theta)}{\sin(\theta)}\)
\(\displaystyle \sin(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{\frac{1}{\sin^2(\theta)}}\)
\(\displaystyle \sin(2\theta)=2\frac{\sin^2(\theta)}{\cos(\theta)}\)
\(\displaystyle \sin(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{\frac{1}{\cos^2(\theta)}}\)
None of the above

How can we simplify the right-hand side?

\(\displaystyle \sin(2\theta)=2\frac{\sin^3(\theta)}{\cos(\theta)}\)
\(\displaystyle \sin(2\theta)=2\frac{1-\sin^2(\theta)}{\sin(\theta)}\)
\(\displaystyle \sin(2\theta)=2\frac{1-\cos^2(\theta)}{\cos(\theta)}\)
\(\displaystyle \sin(2\theta)=2\frac{\sin(\theta)}{\cos(\theta)}\)
\(\displaystyle \sin(2\theta)=\cos^2(\theta)-\sin^2(\theta)\)
\(\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)\)
None of the above

How can we finish the proof?

The previous equation is just the double angle formula for cosine.
The previous equation is just the double angle formula for sine.
Apply the Pythagorean Identity to get that the left and right hand sides are equal.
Multiply by \(\sin(\theta)\) on both sides to then solve for all values of \(\theta\text{.}\)
None of the above

12. Verifying Identities.

Let’s verify the identity \((\csc(\theta)+\cot(\theta))(1-\cos(\theta))=\sin(\theta)\text{.}\) What is a good first step?

\(\displaystyle \csc(\theta)+\cot(\theta)-\csc(\theta)\cos(\theta)-\cot(\theta)\cos(\theta)=\sin(\theta)\)
\(\displaystyle \csc(\theta)-\cot(\theta)+\csc(\theta)\cos(\theta)+\cot(\theta)\cos(\theta)=\sin(\theta)\)
\(\displaystyle \csc(\theta)+\cos(\theta)-\csc(\theta)\cot(\theta)+\cot(\theta)\cos(\theta)=\sin(\theta)\)
\(\displaystyle \csc(\theta)+\cot(\theta)-\cot(\theta)\cos(\theta)=\sin(\theta)\)
\(\displaystyle \csc(\theta)+\cot(\theta)-\csc(\theta)\cos(\theta)+\cot(\theta)\cos(\theta)=\sin(\theta)\)
None of the above

What do we get when we rewrite the left-hand side purely in terms of sine and cosine?

\(\displaystyle \frac{1}{\cos(\theta)}-\frac{\sin(\theta)}{\cos(\theta)}-\frac{\cos(\theta)}{\sin^2(\theta)} -\frac{\cos(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}-\frac{\sin(\theta)}{\cos\theta)} -\frac{\sin^2(\theta)}{\cos(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1}{\sin(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} -\frac{\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1}{\sin(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} +\frac{\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
None of the above

How can we simplify the previous equation?

\(\displaystyle \frac{1-\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{\sin(2\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1-2\cos(\theta)+\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1+\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
None of the above

How can we apply the Pythagorean Identity here?

\(\displaystyle \frac{\sin^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{1+\cos^2(\theta)}{\sin(\theta)}=1-\sin(\theta)\)
\(\displaystyle \frac{1-\cos^2(\theta)}{\sin(\theta)}=1-\sin^2(\theta)\)
\(\displaystyle \frac{\cos^2(\theta)}{\sin(\theta)}=\sin(\theta)\)
\(\displaystyle \frac{\sin(2\theta)}{2\cos(\theta)}=\sin(\theta)\)
None of the above

How can we finish the proof?

After applying the double angle formula, both sides become equal.
After simplifying, we obtain the double angle formula for sine.
After simplifying the left-hand side, both sides are equal.
After applying the Pythagorean Identity again, both sides become equal.
None of the above

13. Verifying Identities.

Let’s verify the identity \(\frac{\sec(\theta)+\csc(\theta)}{1+\tan(\theta)}=\csc(\theta)\text{.}\) What is a good first step?

\(\displaystyle \sec(\theta)+\csc(\theta)=1+\tan(\theta)\csc(\theta)\)
\(\displaystyle \frac{\sec(\theta)+\csc(\theta)}{\sec(\theta)}=\csc(\theta)\)
\(\displaystyle \sec(\theta)+\csc(\theta)=\csc(\theta)+\tan(\theta)\csc(\theta)\)
\(\displaystyle \frac{\sec(\theta)+\csc(\theta)}{\sec^2(\theta)}=\csc(\theta)\)
\(\displaystyle \sec(\theta)+\csc(\theta)=\tan(\theta)\csc(\theta)+\tan(\theta)\csc(\theta)\)
None of the above

What is a quick way to simplify the previous equation?

\(\displaystyle \sec(\theta)+\csc(\theta)=\sec(\theta)\csc(\theta)\)
\(\displaystyle \sec(\theta)=\tan(\theta)\csc(\theta)\)
\(\displaystyle \sec(\theta)+\csc(\theta)=\sec^2(\theta)\csc(\theta)\)
\(\displaystyle \cot(\theta)=2\tan(\theta)\)
\(\displaystyle \cot(\theta)=1+\tan(\theta)\)
None of the above

What do we get when we rewrite the right-hand side purely in terms of sine and cosine?

\(\displaystyle \sec(\theta)=\frac{\sin(\theta)}{\cos^2(\theta)}\)
\(\displaystyle \cot(\theta)=2\frac{\sin(\theta)}{\cos(\theta)}\)
\(\displaystyle \sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\)
\(\displaystyle \sec(\theta)+\csc(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\)
\(\displaystyle \cot(\theta)=1+\frac{\sin(\theta)}{\cos(\theta)}\)
\(\displaystyle \sec(\theta)+\csc(\theta)=\frac{\sin(\theta)}{\cos^2(\theta)}\)
\(\displaystyle \sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)\sin(\theta)}\)
None of the above

How can we finish the proof?

We apply the Pythagorean Identity to make both sides equal.
After simplifying, we get the get double angle formula for sine.
After simplifying the previous equation, we are left with the definition of secant.
We apply the double angle formula for cosine after simplifying.
None of the above

14. Verifying Identities.

Let’s verify the identity \(\cos(2\theta)=\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}\text{.}\) What is a good first step?

\(\displaystyle \cos^2(\theta)-\sin^2(\theta)=\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle 2\sin(\theta)\cos(\theta)=\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle 1-\sin^2(\theta)=\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}\)
None of the above

In this next part, we’ll use a trick: add and subtract \(2\tan^2(\theta)\) to the numerator in the right-hand side. What do you get for your new right-hand side?

\(\displaystyle \frac{1-2\tan^2(\theta)}{1+\tan^2(\theta)}+\frac{+2\tan^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle \frac{1+2\tan^2(\theta)}{1-2\tan^2(\theta)}\)
\(\displaystyle \frac{1+\tan^2(\theta)}{1+\tan^2(\theta)}+\frac{-2\tan^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle \frac{1+\tan^2(\theta)}{1-\tan^2(\theta)}-\frac{2\tan^2(\theta)}{1-\tan^2(\theta)}\)
\(\displaystyle \frac{1+\tan^2(\theta)}{1-\tan^2(\theta)}\)
None of the above

What do we get when we simplify the right-hand side?

\(\displaystyle \frac{1-2\tan^2(\theta)}{1+\tan^2(\theta)}+2\)
\(\displaystyle \frac{\sec^2(\theta)}{1-\tan^2(\theta)}-1\)
\(\displaystyle 1-2\frac{\tan^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle \frac{\sec^2(\theta)}{1+\tan^2(\theta)}\)
\(\displaystyle 2\tan(\theta)\)
None of the above

What do you get when you multiply the fraction in the right-hand side by \(\frac{\cos^2(\theta)}{\cos^2(\theta)}\) and simplify?

\(\displaystyle \frac{\cos^2(\theta)-2\sin^2(\theta)}{\cos^2(\theta)+\sin^2(\theta)}\)
\(\displaystyle 1-2\frac{\sin^2(\theta)}{\cos^2(\theta)+\sin^2(\theta)}\)
\(\displaystyle \frac{1}{\cos^2(\theta)+\sin^2(\theta)}\)
\(\displaystyle 2\frac{\sin^2(\theta)}{\sin(\theta)\cos(\theta)}\)
\(\displaystyle \frac{\cos^2(\theta)}{\cos^2(\theta)-\sin^2(\theta)}\)
None of the above

Let’s simplify the previous expression and figure out what equation we are working with now. What do we have?

\(\displaystyle 2\sin(\theta)\cos(\theta)=1-2\sin^2(\theta)\)
\(\displaystyle \cos^2(\theta)-\sin^2(\theta)=\sec^2(\theta)+\sin^2(\theta)\)
\(\displaystyle 2\sin(\theta)\cos(\theta)=\cos^2(\theta)\)
\(\displaystyle \cos^2(\theta)-\sin^2(\theta)=1-2\sin^2(\theta)\)
\(\displaystyle \cos^2(\theta)-\sin^2(\theta)=2\tan(\theta)\)
\(\displaystyle 1-\sin^2(\theta)=2\tan(\theta)\)
None of the above

How can we finish the proof?

After rearranging terms and simplifying, we are left with the definition of secant.
We may add \(2\sin^2(\theta)\) to both sides to obtain the Pythagorean Identity.
After rearranging terms, both sides are identical.
After rearranging terms and simplifying, we are left with the double angle formula for sine.
None of the above

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