Hi Everyone!

On this page you will find some material about Lesson 30. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 30: Fundamental Identities & Proving Trigonometric Tautologies**

**Learning Outcomes.** (from Coburn and Herdlick’s Trigonometry book)

- Use fundamental identities to help understand and recognize identity “families.”
- Verify other identities using the fundamental identities and basic algebra skills.
- Use fundamental identities to express a given trig function in terms of the other five.
- Identify and use identities due to symmetry.
- Verify general identities.
- Use counterexamples and contradictions to show an equation is not an identity.

**Topic**. This lesson covers

Section 1.4: Fundamental Identities and Families of Identities, and

Section 5.1: More on Verifying identities.

**WeBWorK**. There is no WeBWorK assignment on today’s material, but Lesson 30 Notes – part 2 have some homework.

**Lesson Notes.**

**Video Lesson.**

Video Lesson 30 (based on Lesson 30 Notes – part 1)

### Warmup Questions

*These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.*

#### Warmup Question 1

The point $(\cos x,\sin x)$ belongs to the unit circle, so it must satisfy the equation of the unit circle. What does that mean?

#### Show Answer 1

The equation of the unit circle is $x^2+y^2=1$. If $(\cos x,\sin x)$ belongs to the unit circle, then we must have

$$\cos^2x +\sin^2 x=1.$$

This is one of the most fundamental trigonometric identities.

#### Warmup Question 2

Can you write $\sec x$, $\csc x$ and $\cot x$ in terms of $\cos x$, $\sin x$ or $\tan x$?

#### Show Answer 2

$$\sec x=\dfrac{1}{\cos x}$$

$$\csc x=\dfrac{1}{\sin x}$$

$$\tan x=\dfrac{1}{\tan x}$$

#### Warmup Question 3

Subtract $\dfrac{b}{a}-\dfrac{a}{b}$.

#### Show Answer 3

$$\dfrac{a}{b}-\dfrac{b}{a}=\dfrac{a^2}{ab}-\dfrac{b^2}{ab} =\dfrac{a^2-b^2}{ab}$$

### Review

*If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.*

Need a review? Check

### Quick Intro

*This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.*

**A Quick Intro to Fundamental Identities and Families of Identities**

**Key Words.** Fundamental Identities

$\bigstar$ Recall the Fundamental Identities:

$$\sin^2\theta +\cos^2\theta =1$$

$$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$$

$$\csc\theta=\dfrac{1}{\sin\theta}$$

$$\sec\theta=\dfrac{1}{\cos\theta}$$

$$\cot\theta=\dfrac{1}{\tan\theta}$$

$\bigstar$ Dividing $\sin^2\theta+\cos^2\theta = 1$ by $\sin^2\theta$ gives:

$$1+\cot^2\theta = \csc^2\theta.$$

$\bigstar$ Dividing $\sin^2\theta+\cos^2\theta = 1$ by $\cos^2\theta$ gives:

$$\tan^2\theta +1 = \sec^2\theta.$$

Check the following notes for a simple method to show trig identities.

### Video Lesson

*Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!*

**Video Lesson**

**A description of the video**

In the video you will see how to show

- $\tan\theta\sin\theta + \cos\theta = \sec\theta$
- $\dfrac{\sin x}{1-\cos x} =\csc x-\cot x$

### Try Questions

*Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.*

#### Try Question 1

Show that $\sec^2\theta\cot^2\theta = \csc^2\theta$.

#### Show Answer 1

$$\sec^2\theta\cot^2\theta \stackrel{?}{=} \csc^2\theta$$

$$\dfrac{1}{\cos^2\theta}\dfrac{\cos^2\theta}{\sin^2\theta}\stackrel{?}{=} \dfrac{1}{\sin^2\theta}$$

Let $a=\cos\theta$ and $b=\sin\theta$.

$$\dfrac{1}{a^2}\dfrac{a^2}{b^2}\stackrel{?}{=} \dfrac{1}{b^2}$$

$$\dfrac{1}{b^2}\stackrel{\checkmark}{=} \dfrac{1}{b^2}$$

#### Try Question 2

Show $\cos\theta(\sec\theta-\cos\theta)=\sin^2\theta$.

#### Show Answer 2

$$\cos\theta(\sec\theta-\cos\theta)\stackrel{?}{=}\sin^2\theta$$

$$\cos\theta\left(\dfrac{1}{\cos\theta}-\cos\theta\right)\stackrel{?}{=}\sin^2\theta$$

Let $a=\cos\theta$ and $b=\sin\theta$.

$$a\left(\dfrac{1}{a}-a\right)\stackrel{?}{=}b^2$$

$$a\left(\dfrac{1}{a}-\dfrac{a^2}{a}\right)\stackrel{?}{=}b^2$$

$$a\left(\dfrac{1-a^2}{a}\right)\stackrel{?}{=}b^2$$

$$1-a^2\stackrel{?}{=}b^2$$

$$1\stackrel{\checkmark}{=}a^2+b^2$$

### WeBWorK

*You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.*

**WeBWork**

There is no homework on WeBWork for this lesson, but the following lecture notes have a list of problems on page 7.

When you are done, come back to this page for the Exit Questions.

### Exit Questions

*After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!*

#### Exit Questions

- Does it make sense to solve a trig identity?
- What are key basic identities that you might use?

$\bigstar$ Prove that $\dfrac{\sec^2x}{1+\cot^2x} =\tan^2x$

#### Show Answer

$$\dfrac{\sec^2x}{1+\cot^2x}\stackrel{?}{=} \tan^2x$$

$$\dfrac{(\sec x)^2}{1+(\cot x)^2}\stackrel{?}{=}(\tan x)^2$$

$$\dfrac{\left(\dfrac{1}{\cos x}\right)^2}{1+\left(\dfrac{\cos x}{\sin x}\right)^2} \stackrel{?}{=}\left(\dfrac{\sin x}{\cos x}\right)^2$$

$$\dfrac{\left(\dfrac{1}{a}\right)^2}{1+\left(\dfrac{a}{b}\right)^2}\stackrel{?}{=}\left(\dfrac{b}{a}\right)^2$$

Let $a=\cos x$ and $b=\sin x$.

$$\dfrac{\dfrac{1}{a^2}}{1+\dfrac{a^2}{b^2}}\stackrel{?}{=}\dfrac{b^2}{a^2}$$

$$\dfrac{\dfrac{1}{a^2}}{\dfrac{b^2}{b^2}+\dfrac{a^2}{b^2}}\stackrel{?}{=}\dfrac{b^2}{a^2}$$

$$\dfrac{\dfrac{1}{a^2}}{\dfrac{b^2+a^2}{b^2}}\stackrel{?}{=}\dfrac{b^2}{a^2}$$

Recall: $a^2+b^2=1$.

$$\dfrac{\dfrac{1}{a^2}}{\dfrac{1}{b^2}}\stackrel{?}{=}\dfrac{b^2}{a^2}$$

$$\dfrac{1}{a^2}\cdot \dfrac{b^2}{1}\stackrel{?}{=}\dfrac{b^2}{a^2}$$

$$\dfrac{b^2}{a^2} \stackrel{\checkmark}{=}\dfrac{b^2}{a^2}$$

### Need more help?

*Don’t wait too long to do the following.*

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.