Section 6.1: Fundamental Trigonometric Identities
Learning Objectives
In this section you will:
- Verify the fundamental trigonometric identities.
- Simplify trigonometric expressions using algebra and the identities.
Introduction
In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.
The Reciprocal and Quotient Trigonometric Identities
We'll start with the familiar: the reciprocal identities. We encountered these way back in Section 4.1 when we first learned right triangle trigonometry.
The Reciprocal Identities
Show that $\sin(\theta) = \frac{1}{\csc(\theta)}$.
We can work from the definition of $\csc(\theta)$, which is $\csc(\theta) = \frac{1}{\sin(\theta)}$:
Multiply by sine: $$\sin(\theta)\cdot \csc(\theta) = 1$$ Divide by $\csc(\theta)$: $$\sin(\theta) = \frac{1}{\csc(\theta)}$$This is where all the identities on the right hand column are coming from.
When we learned about the unit circle, we also discovered a different way of defining tangent and cotangent; these are known as the quotient identities.
The Quotient Identities
The Pythagorean Identities
Many other identities can be discovered by using the physical properties of the Unit Circle. To build up our Unit Circle, we first thought about embedded right triangles within it:
When we first built the unit circle, we found that $x = \cos(\theta)$ and $y = \sin(\theta)$. Now, consider the implications for the Pythagorean Theorem: $$x^2 + y^2 = 1^2$$ $$(\cos(\theta))^2 + (\sin(\theta))^2 = 1$$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$ The magic is, this is a relationship that holds true no matter the value of $\theta$! This is known as the Pythagorean Identity. Using $$\cos^2(\theta) + \sin^2(\theta) = 1$$ as a starting point, we can then come up with a couple of versions of the Pythagorean Identity.
\[\begin{align*} (\cos^2(\theta))~~ + (\sin^2(\theta)) &= ~~1 &\textrm{}\\ ~~&~~\\ \underline{\mathstrut{\div \cos^2(\theta) ~~~ \div \cos^2(\theta) }}&\underline{\mathstrut{~~\div \cos^2(\theta)}}&\textrm{ Divide $\cos^2(\theta)$ from both sides of the equation.}\\ ~~&~~\\ 1 ~~~~ + \frac{\sin^2(\theta)}{\cos^2(\theta)} &= ~~\frac{1}{\cos^2(\theta)} & \textrm{}\\ \end{align*}\] Now we consider our reciprocal and quotient identities. Since \[\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)\] we can square both sides of this identity to see that \[\frac{\sin^2(\theta)}{\cos^2(\theta)} = \tan^2(\theta).\] Similarly, $\frac{1}{\cos^2(\theta)} = \sec^2(\theta)$. Therefore, our new version of the Pythagorean Identity is \[1 + \tan^2(\theta) = \sec^2(\theta)\]
Create another version of the Pythagorean identity by dividing $\cos^2(\theta) + \sin^2(\theta) = 1$ by $\sin^2(\theta)$
In summary, here are all three of the Pythagorean Identities. They are worth memorizing, although if you know the first, you can always recreate the latter two using the methods from above.
The Pythagorean Identities
For all $\theta$ in the domain...
$$\cos^2(\theta) + \sin^2(\theta) = 1$$ $$1 + \tan^2(\theta) = \sec^2(\theta)$$ $$\cot^2(\theta) + 1 = \csc^2(\theta)$$
The Even/Odd Identities
A function is called even if it has y-axis symmetry; a function is called odd if it has origin symmetry. In Chapter 0, we learned algebraic ways of detecting this type of symmetry:
Testing the Graph of an Equation for Symmetry
To test the graph of an equation for symmetry
- About the $y$-axis: Substitute $(-x,y)$ into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the $y$-axis.
- About the origin: Substitute $(-x,-y)$ into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the origin.
We can modify these notions to think about functions instead of merely equations. For instance, for $y$-axis symmetry, let $y = f(x)$. So if we substitute $-x$ into our function, we want the result to be equivalent to the original function; that is $f(-x) = f(x)$. If a function has this property, we will call it even. Similarly, for origin symmetry, let $y = f(x)$. If we substitute in $-x$, we should get back $-y = -f(x)$; that is, $f(-x) = -f(x)$. If a function has this property, we will call it odd.
Even and Odd Functions
A function is even if its graph has $y$-axis symmetry. Then the function will also have the property that $f(-x) = f(x)$.
A function is odd if its graph has origin symmetry. Then the function will also have the property that $f(-x) = -f(x)$.
Now, consider our trigonometric functions. The graph of $f(x) = \cos(x)$ is symmetric about the $y$-axis, so it is even. More relevant to us now, that means that $\cos(-x) = \cos(x)$. The graph of $f(x) = \sin(x)$ is symmetric about the origin, so it is odd and $\sin(-x) = -\sin(x)$. From there, we can derive the rest of the even/odd identities.
The Even/Odd Identities
$\cos(\theta)$ and $\sec(\theta)$ are even, so for all $\theta$ in the domain...
$$\cos(-\theta) = \cos(\theta)$$
$$\sec(-\theta) = \sec(\theta).$$
The rest of the trigonometric functions are odd, hence for all $\theta$ in the domain...
$$\sin(-\theta) = -\sin(\theta)$$
$$\csc(-\theta) = -\csc(\theta)$$
$$\tan(-\theta) = -\tan(\theta)$$
$$\cot(-\theta) = -\cot(\theta)$$
One of the most important things that we will use these identities for is simplification. As you move on to other math class (like Calculus in particular), the goals are much less "solve for $x$" and much more "manipulate this expression so it's more convenient."
When we simplify trig expressions, we're generally trying to write the whole thing in terms of a single trigonometric function. There's not always a single path forward to this goal; a good way to practice your simplification skills is to try to approach a particular problem from a few different points of view. The best way to get good at this skill is through experience and practice!
$$\sin(x)\cos(-x)\sec(x)$$
First, we want everything to be in terms of the same angle. Currently, sine and secant have a positive $x$ input, but cosine has $-x$. We apply the even identity for cosine:
\[\begin{array}{c |c}
\text{Expression} & \text{Reasoning}\\
&\\
\sin(x)\textcolor{blue}{\cos(-x)}\sec(x) & \\
&\\
=\sin(x)\textcolor{blue}{\cos(x)}\sec(x) & \text{$\cos(x)$ is even, so $\cos(-x) = \cos(x)$}\\
&\\
\end{array}\]
Now, we have a mix of different trigonometric funtions. In such cases, it is often a good idea to convert everything to sines and cosines. Here we can use the reciprocal identities to help:
\[\begin{array}{c |c}
\text{Expression} & \text{Reasoning}\\
&\\
=\sin(x)\cos(x)\textcolor{blue}{\sec(x)} & \\
&\\
=\sin(x)\cos(x)\textcolor{blue}{\frac{1}{\cos(x)}} & \text{Reciprocal identity: $\sec(x) = \frac{1}{\cos(x)}$}\\
\end{array}\]
Finally, we use algebra to simplify:
\[\begin{array}{c|c}
\text{Expression} & \text{Reasoning}\\
&\\
=\sin(x)\textcolor{blue}{\cos(x)\cdot\frac{1}{\cos(x)}} &\\
&\\
=\sin(x)\textcolor{blue}{\frac{\cos(x)}{\cos(x)}} & \text{Algebra}\\
&\\
=\sin(x) \textcolor{blue}{\cdot 1} & \text{Algebra}\\
&\\
=\sin(x)&\\
\end{array}\]
Therefore, the expression $\sin(x)\cos(-x)\sec(x)$ simplifies to $\sin(x)$.
$$\frac{\cot(t) + \tan(t)}{\sec(-t)}$$
Hence, $\frac{\cot(t) + \tan(t)}{\sec(-t)}$ simplifies to $\csc(t)$.
Using Algebra and Identities to Simplify Trigonometric Expressions
We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.
Conider, for example, the difference of squares formula, $a^2−b^2=(a−b)(a+b)$. We often make use of this formula when simplifying expressions like $\cos^4(x) - 1$, which can be factored as $(\cos^2(x)+1)(\cos^2(x)-1)$. We can use many such factoring strategies when simplifying trigonometric expressions.
$$\frac{\sin^4(x) - 1}{\sin^2(x)+1}$$
Now, when we see a squared trig function and a 1, that should put the Pythagorean identities in mind. Consider the first identity; since it is true for all values of $\theta$ in the domain, we can rearrange it as we please: $$\begin{align*} \cos^2(x) + \sin^2(x) &= 1\\ \underline{\mathstrut{-\cos^2(x)~~~~~~~~~~~~~}}&\underline{\mathstrut{~-\cos^2(x)}}\\ \sin^2(x) &= 1 - \cos^2(x)\\ \underline{\mathstrut{-1}} & \underline{\mathstrut{~~~-1 ~~~~~~}}\\ \sin^2(x) - 1 &= -\cos^2(x) \end{align*}$$
Hence our expression above, $\sin^2(x) - 1$ can be simplified using the Pythagorean Identity to $-\cos^2(x)$.
We could have approached this more algebraically as well. Since the Pythagorean Identity states the $\cos^2(x) + \sin^2(x) = 1$, we could replace the $1$ in our expression with $\cos^2(x) + \sin^2(x)$: \[\begin{array}{c | c} \text{Expression} & \text{Reasoning}\\ =\sin^2(x) - \textcolor{blue}{1} & \\ &\\ =\sin^2(x) - \textcolor{blue}{(\cos^2(x) + \sin^2(x)) }& \text{Pythagorean Identity}\\ &\\ =\sin^2(x) - \cos^2(x) - \sin^2(x) & \text{Distribute the negative}\\ &\\ =\cancel{\sin^2(x)} - \cos^2(x) - \cancel{\sin^2(x)} & \text{$\sin^2(x) - \sin^2(x)$ is 0}\\ &\\ =-\cos^2(x) & \\ \end{array}\]
Either way, $\frac{\sin^4(x) - 1}{\sin^2(x)+1}$ simplifies to $-\cos^2(x)$.
Verifying Trigonometric Identities
In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. To be mathematically rigorous, we must commit to using only one side. Strategies such as multiplying by the same quantity on both sides only work in the context of solving an equation, and rely on the assumption that the two sides of the equation will remain balanced. Here our goal is different: we are trying to prove the two sides of the equation are exactly the same for all values of the variable in the domain. So, overall we can think of this process as simplifying towards a specific goal.
$$\tan(\theta) = \sin(\theta)\sec(\theta)$$
Hence we have shown that the left hand side is equal to the right hand side, so we have verified the identity holds for all values of $\theta$.
$$(\tan(t) - \sec(t)) (\tan(t) + \sec(t)) = -1$$
$$\dfrac{\sec(t)}{1 - \tan(t)} = \dfrac{1}{\cos(t) - \sin(t)}$$
$$\dfrac{\sin(\theta)}{1 - \cos(\theta)} = \dfrac{1 + \cos(\theta)}{\sin(\theta)}$$
In the example above, we see that multiplying $1-\cos(\theta)$ by $1+\cos(\theta)$ produces a difference of squares that can be simplified to one term using the Pythagorean Identity.
This is exactly the same kind of phenomenon that occurs when we multiply expressions such as $1 - \sqrt{2}$ by $1+\sqrt{2}$ or $3 - 4i$ by $3+4i$. In algebra, these sorts of expressions were called `conjugates.'
Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proficient in the basics.
Like many things in life, there is no short-cut here -- there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises.
Strategies for Verifying Identities
- Try working on the more complicated side of the identity.
- Use the Reciprocal and Quotient Identities in Theorem to write functions on one side of the identity in terms of the functions on the other side of the identity.
Simplify the resulting complex fractions.
- Add rational expressions with unlike denominators by obtaining common denominators.
- Use the Pythagorean Identities in Theorem to 'exchange' sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term.
- Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities.
- If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work.
- Try something. Do algebra and have faith, even if you can't see the path forward. The more you work with identities, the better you'll get with identities.
Practice Problems
Verify each identity.1. $\cos(\theta) \sec(\theta) = 1$
2. $\tan(t)\cos(t) = \sin(t)$
3. $\sin(\theta) \csc(\theta) = 1$
4. $\tan(t) \cot(t) = 1$
5. $\csc(x) \cos(x) = \cot(x)$
6. $\dfrac{\sin(t)}{\cos^{2}(t)} = \sec(t) \tan(t)$
7. $\dfrac{\cos(\theta)}{\sin^{2}(\theta)} = \csc(\theta) \cot(\theta)$
8. $\dfrac{1+ \sin(x)}{\cos(x)} = \sec(x) + \tan(x)$
9. $\dfrac{1 - \cos(\theta)}{\sin(\theta)} = \csc(\theta) - \cot(\theta)$
10. $\dfrac{\cos(t)}{1 - \sin^{2}(t)} = \sec(t)$
11. $\dfrac{\sin(x)}{1 - \cos^{2}(x)} = \csc(x)$
12. $\dfrac{\sec(t)}{1 + \tan^{2}(t)} = \cos(t)$
13. $\dfrac{\csc(\theta)}{1 + \cot^{2}(\theta)} = \sin(\theta)$
14. $\dfrac{\tan(x)}{\sec^{2}(x) - 1} = \cot(x)$
15. $\dfrac{\cot(t)}{\csc^{2}(t) - 1} = \tan(t)$
16. $4 \cos^{2}(\theta) + 4 \sin^{2}(\theta) = 4$
17. $9 - \cos^{2}(t) - \sin^{2}(t) = 8$
18. $\tan^{3}(t) = \tan(t)\sec^{2}(t) - \tan(t)$
19. $\sin^{5}(x) = \left(1-\cos^{2}(x)\right)^{2} \sin(x)$
20. $\sec^{10}(t) = \left(1 + \tan^{2}(t)\right)^4 \sec^{2}(t)$
21. $\cos^{2}(x)\tan^{3}(x) = \tan(x) - \sin(x)\cos(x)$
22. $\sec^{4}(t) - \sec^{2}(t) = \tan^{2}(t) + \tan^{4}(t)$
23. $\dfrac{\cos(\theta) + 1}{\cos(\theta) - 1} = \dfrac{1 + \sec(\theta)}{1 - \sec(\theta)}$
24. $\dfrac{\sin(t) + 1}{\sin(t) - 1} = \dfrac{1 + \csc(t)}{1 - \csc(t)}$
25. $\dfrac{1 - \cot(x)}{1+ \cot(x)} = \dfrac{\tan(x) - 1}{\tan(x) + 1}$
26. $\dfrac{1 - \tan(t)}{1+ \tan(t)} = \dfrac{\cos(t) - \sin(t)}{\cos(t) + \sin(t)}$
27. $\tan(\theta) + \cot(\theta) = \sec(\theta)\csc(\theta)$
28. $\csc(t) - \sin(t) = \cot(t)\cos(t)$
29. $\cos(x) - \sec(x) = -\tan(x)\sin(x)$
30. $\cos(x)(\tan(x) + \cot(x)) = \csc(x)$
31. $\sin(t)(\tan(t) + \cot(t)) = \sec(t)$
32. $\dfrac{1}{1-\cos(\theta)} + \dfrac{1}{1+\cos(\theta)} = 2\csc^{2}(\theta)$
33. $\dfrac{1}{\sec(t) + 1} + \dfrac{1}{\sec(t)-1} = 2 \csc(t) \cot(t)$
34. $\dfrac{1}{\csc(x) + 1} + \dfrac{1}{\csc(x)-1} = 2 \sec(x) \tan(x)$
35. $\dfrac{1}{\csc(t)-\cot(t)} - \dfrac{1}{\csc(t) + \cot(t)} = 2 \cot(t)$
36. $\dfrac{\cos(\theta)}{1 - \tan(\theta)} + \dfrac{\sin(\theta)}{1 - \cot(\theta)} = \sin(\theta) + \cos(\theta)$
37. $\dfrac{1}{\sec(t) + \tan(t)} = \sec(t) - \tan(t)$
38. $\dfrac{1}{\sec(x) - \tan(x)} = \sec(x) + \tan(x)$
39. $\dfrac{1}{1-\cos(\theta)} = \csc^{2}(\theta) + \csc(\theta) \cot(\theta)$
40. $\dfrac{1}{1+\cos(x)} = \csc^{2}(x) - \csc(x) \cot(x)$
Homework Set
Simplify each expression
- $\sin(-x)\cos(-x)\csc(x)$
- $\cot(x)\cos(x)+\csc(x)\sin^2(x)$
- $\csc(x)+\cos(x)\cot(-x)$
- $\frac{\tan(t)+\cot(t)}{\csc(-t)}$
- $3\sin^3(t)\csc(t)+\cos^2(t)+2\cos(-t)\cos(t)$
- $\frac{1+\cot^2(x)}{\sec^2(x)}+\cos^2(x)+\frac{1}{\csc^2(x)}$
- $\frac{\cos^4(x)-1}{\cos^2(x)+1}$
- $\cos(x)(\tan(x)-\sec(-x))=\sin(x)-1$
- $(\sin(x)+\cos(x))^2=1+2\sin(x)\cos(x)$
- $\frac{1+\sin^2(x)}{\cos^2(x)} = 1+2\tan^2(x)$
- $\frac{1}{\sec(t)+\tan(t)}=\sec(t)-\tan(t)$
- $\cos^2(x)\tan^3(x)=\tan(x)-\sin(x)\cos(x)$