Trig Identities
Instructions
- The first videos below explain the concepts in this section.
- This page includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
- When you are done, you can use the graph to pick another section or use the buttons to go to the next section.
Learning Objectives
- Introducing the Basic 8 trigonometric identities: the reciprocal, ratio, and Pythagorean identities
- Using trigonometric identities to simplify trigonometric expressions
- Deriving trig identities from existing identities
Concept Video(s)
Exercises
1.
Answer: \(\tan x\)
Solution Method:The Reciprocal and Ratio Identities are the most basic identities. The three reciprocal identities simply define the reciprocal functions. Then the ratio identities define tangent and cotangent in terms of sine and cosine.
\csc{x}=\dfrac{1}{\sin{x}} & \tan{x} = \dfrac{\sin{x}}{\cos{x}} \\[8pt]
\sec{x}=\dfrac{1}{\cos{x}} & \cot{x} = \dfrac{\cos{x}}{\sin{x}}\\[8pt]
\cot{x}=\dfrac{1}{\tan{x}}
\end{array}
When simplifying an equation using trig identities, our goal is to get to a single function or number. A helpful trick is to get everything in terms of sine and cosine, because all trig functions can be defined by those two, as you see from these identities.
\begin{equation*}
\frac{\sin^2{x}}{\cos{x}} \cdot \csc{x} =
\frac{\sin^2{x}}{\cos{x}} \cdot \frac{1}{\sin{x}} =
\frac{\sin{x}}{\cos{x}} =
\tan{x}
\end{equation*}
Solution Method:The Reciprocal and Ratio Identities are the most basic identities. The three reciprocal identities simply define the reciprocal functions. Then the ratio identities define tangent and cotangent in terms of sine and cosine.
Reciprocal and Ratio Identities
\begin{array}{cc}\csc{x}=\dfrac{1}{\sin{x}} & \tan{x} = \dfrac{\sin{x}}{\cos{x}} \\[8pt]
\sec{x}=\dfrac{1}{\cos{x}} & \cot{x} = \dfrac{\cos{x}}{\sin{x}}\\[8pt]
\cot{x}=\dfrac{1}{\tan{x}}
\end{array}
When simplifying an equation using trig identities, our goal is to get to a single function or number. A helpful trick is to get everything in terms of sine and cosine, because all trig functions can be defined by those two, as you see from these identities.
\begin{equation*}
\frac{\sin^2{x}}{\cos{x}} \cdot \csc{x} =
\frac{\sin^2{x}}{\cos{x}} \cdot \frac{1}{\sin{x}} =
\frac{\sin{x}}{\cos{x}} =
\tan{x}
\end{equation*}
2.
Answer: 1
Solution Method: The primary Pythagorean identity \(\sin^2{x} + \cos^2{x} = 1\) comes from the application of the Pythagorean theorem on the unit circle. On the unit circle \(x=\text{cosine}\) and \(y=\text{sine},\) and the radius is 1. So \begin{align*}
x^2+y^2 &= 1 \\[6pt]
\cos^2{x}+\sin^2{x} &= 1
\end{align*}
The other 2 Pythagorean identities are derived from the first using the Reciprocal and Ratio Identities.
\sin^2{x} + \cos^2{x} = 1 \\[6pt]
1 + \tan^2{x} = \sec^2{x} \\[6pt]
1 + \cot^2{x} = \csc^2{x}
\end{gather}
We want to simplify \((\sec{x}-\tan{x})(\sec{x}+\tan{x})\) as much as possible, hopefully to a single trig function or number. First, this is a difference of squares so multiplied out it is
\begin{align*}
(\sec{x}-\tan{x})(\sec{x}+\tan{x}) = \sec^2{x}-\tan^2{x}
\end{align*}
It's fairly clear that we're going to use the identity \(1 + \tan^2{x} = \sec^2{x}\) to simplify this further. So I substitute \(\sec^2{x}\) with \(1 + \tan^2{x}\),
\begin{align*}
\sec^2{x}-\tan^2{x} &= 1+ \tan^2{x} -\tan^2{x} \\
&= 1
\end{align*}
Therefore, I have fully simplified our original expression
\[
(\sec{x}-\tan{x})(\sec{x}+\tan{x})=1
\]
Notice I also could have rearranged the identity to match the RHS,
\begin{align*}
1 + \tan^2{x} = \sec^2{x} \\
1 = \sec^2{x} - \tan^2{x}
\end{align*}
and reached the same answer that way.
Solution Method: The primary Pythagorean identity \(\sin^2{x} + \cos^2{x} = 1\) comes from the application of the Pythagorean theorem on the unit circle. On the unit circle \(x=\text{cosine}\) and \(y=\text{sine},\) and the radius is 1. So \begin{align*}
x^2+y^2 &= 1 \\[6pt]
\cos^2{x}+\sin^2{x} &= 1
\end{align*}
The other 2 Pythagorean identities are derived from the first using the Reciprocal and Ratio Identities.
Pythagorean Identities
\begin{gather}\sin^2{x} + \cos^2{x} = 1 \\[6pt]
1 + \tan^2{x} = \sec^2{x} \\[6pt]
1 + \cot^2{x} = \csc^2{x}
\end{gather}
We want to simplify \((\sec{x}-\tan{x})(\sec{x}+\tan{x})\) as much as possible, hopefully to a single trig function or number. First, this is a difference of squares so multiplied out it is
\begin{align*}
(\sec{x}-\tan{x})(\sec{x}+\tan{x}) = \sec^2{x}-\tan^2{x}
\end{align*}
It's fairly clear that we're going to use the identity \(1 + \tan^2{x} = \sec^2{x}\) to simplify this further. So I substitute \(\sec^2{x}\) with \(1 + \tan^2{x}\),
\begin{align*}
\sec^2{x}-\tan^2{x} &= 1+ \tan^2{x} -\tan^2{x} \\
&= 1
\end{align*}
Therefore, I have fully simplified our original expression
\[
(\sec{x}-\tan{x})(\sec{x}+\tan{x})=1
\]
Notice I also could have rearranged the identity to match the RHS,
\begin{align*}
1 + \tan^2{x} = \sec^2{x} \\
1 = \sec^2{x} - \tan^2{x}
\end{align*}
and reached the same answer that way.