## Pre-Calculus For Dummies, 2nd Edition (2012)

### Part II. The Essentials of Trigonometry

### Chapter 8. Identifying with Trig Identities: The Basics

*In This Chapter*

Reviewing the basics of solving trig equations

Simplifying and proving expressions with fundamental trig identities

Handling more complicated proofs

In this chapter and the next, you work on simplifying expressions using basic trig identities to prove more complicated identities and solving equations that involve trig functions. Because we’re fans of building momentum, we’re going to start off slow with the basics. This chapter covers basic *identities,* which are statements that are always true and that you use throughout an entire equation to help you simplify the problem, one expression at a time, prior to solving it. And you’re in luck, because trig has many of these identities.

The hard thing about simplifying trig expressions by using trig identities, though, is knowing when to stop. Enter *proofs,* which give you an end goal so that you know when you’ve hit a stopping point. You use proofs when you need to show that two expressions are equal even though they look completely different. (If you thought you were done with proofs when you moved past geometry, think again!)

If you listen to one piece of advice, heed this: Know these basic identities forward, backward, and upside down, because you’ll use them consistently in pre-calculus, and they’ll make your life easier when things get complicated. If you commit the basic identities to memory but can’t remember the more complex identities we present in Chapter 9 (which is highly likely), you can simply use a combination of the basic identities to derive a new identity that suits your situation. This chapter shows you the way.

This chapter focuses on two main ideas that are central to your studies in pre-calc, simplifying expressions and proving complicated identities. These concepts share one common theme: They both involve the trig functions you’re introduced to back in Chapter 6.

**Keeping the End in Mind: A Quick Primer on Identities**

The road to get to the end of each type of problem in this chapter is very similar. However, the final results of the two main ideas — simplifying expressions and proving complicated identities — are different, and that’s what we discuss in this section:

**Simplifying:** Simplifying algebraic expressions with real numbers and variables is nothing new to you (we hope); you base your simplification process on properties of algebra and real numbers that you *know* are always true. In this chapter, we give you some new rules to play by so that you can work with trig problems of all different sorts. Think of basic pre-calc identities as tools in your toolbox to help you build your trig house. One tool on its own doesn’t do you any good. However, when you put the identities together, all the things you can do with them may surprise you! (For instance, you can make a trig expression with many different functions simplify into one function, or perhaps simplify to be real numbers [usually 0 or 1].)

**Proving:** Trig proofs have equations with two sides, and your job is to make one side look like the other. Usually, you make the more-complicated side look like the less-complicated side. Sometimes, though, if you can’t get one side to look like the other, you can “cheat” and work on the other side for a while (or both at the same time). In the end, though, you need to show that one side transforms into the other.

**Lining Up the Means to the End: Basic Trig Identities**

If you’re reading this book straight through, you’ll likely recognize some of the identities in this chapter, because we brush over them in earlier chapters. We use trig functions in Chapter 6 when we examine the ratios between the sides of right triangles. By definition, the reciprocal trig functions are identities, because they’re true for all angle values. We reserved the full discussion of identities for this chapter because the identities aren’t necessary to do the mathematical calculations in the earlier chapters. Now, however, you’re ready to expand your horizons and work with some more-complicated (yet still basic) identities.

In the following sections, we introduce you to the most basic (and most useful) identities. With this information, you can manipulate complicated trig expressions into expressions that are much simpler and more user friendly. This simplification process is one that takes a lot of practice. However, after you master simplifying trig expressions, proving complex identities and solving complicated equations is a breeze.

If each step you take to simplify, prove, or solve a problem with trig is based on an identity (and executed correctly), you’re pretty much guaranteed to get the right answer; the particular route you take to get there doesn’t matter. However, fundamental math skills still apply; you can’t just throw out math rules willy-nilly. Following are some important, fundamental rules that people often forget when working with identities:

Dividing a fraction by another fraction is the same as multiplying by its reciprocal.

To add or subtract two fractions, you must find the common denominator.

You should always factor out the greatest common factor and factor trinomials (see Chapter 4).

**Reciprocal identities**

When you’re asked to simplify an expression involving cosecant, secant, or cotangent, you change the expression to functions that involve sine, cosine, or tangent, respectively. You do this step so that you can cancel functions and simplify the problem. When you change functions in this manner, you’re using the *reciprocal* *identities.* (Technically, the identities are trig functions that just happen to be considered identities as well because they help you simplify expressions.)

The following list presents these reciprocal identities:

Every trig ratio can be written as a combination of sines and/or cosines, so changing all the functions in an equation to sines and cosines is the simplifying strategy that works most often. Always try to do this step first and then look to see if things cancel and simplify. Also, dealing with sines and cosines is usually easier if you’re looking for a common denominator for fractions. From there, you can use what you know about fractions to simplify as much as you can.

**Simplifying an expression with reciprocal identities**

Look for opportunities to use reciprocal identities whenever the problem you’re given contains secant, cosecant, or cotangent. All these functions can be written in terms of sine and cosine, and sines and cosines are always the best place to start. For example, you can use reciprocal identities to simplify this expression:

Follow these steps:

**1. Change all the functions into versions of the sine and cosine functions.**

Because this problem involves a cosecant and a cotangent, you use the reciprocal identities to change

and

This process gives you

**2. Break up the complex fraction by rewriting the division bar that’s present in the original problem as ÷.**

**3. Invert the last fraction and multiply.**

**4. Cancel the functions to simplify.**

The sines and cosines cancel, and you end up getting 1 as your answer.

**Working backward: Using reciprocal identities to prove equalities**

Math teachers frequently ask you to prove complicated identities, because the process of proving those identities helps you to wrap your brain around the conceptual side of math. Oftentimes you’ll be asked to prove identities that involve the secant, cosecant, or cotangent functions. Whenever you see these functions in a proof, the reciprocal identities usually are the best places to start. Without the reciprocal identities, you can go in circles all day without ever actually getting anywhere.

For example, to prove tan *θ* · csc *θ* = sec *θ*, you can work with the left side of the equality only. Follow these simple steps:

**1.** **Convert all functions to sines and cosines.**

The left side of the equation now looks like this:

**2.** **Cancel all possible terms.**

Canceling gives you

which simplifies to

**3.** **You can’t leave the reciprocal function in the equality, so convert back again.**

Because , this equation becomes sec *θ* = sec *θ*. Bingo!

**Pythagorean identities**

The *Pythagorean identities* are among the most useful identities because they simplify complicated expressions so nicely. When you see a trig function that’s squared (sin^{2}, cos^{2}, and so on), keep these identities in mind. They’re built from previous knowledge of right triangles and the alternate trig function values (which we explain in Chapter 6). Recall that the *x-*leg is cos *θ*, the *y-*leg is sin *θ*, and the hypotenuse is 1. Because you know that leg^{2} + leg^{2} = hypotenuse^{2}, thanks to the Pythagorean theorem, then you also know that sin^{2 }*θ* + cos^{2 }*θ* = 1. In this section, we show you where these important identities come from and then how to use them.

The three Pythagorean identities are

sin^{2 }*x* + cos^{2 }*x* = 1

1 + cot^{2 }*x* = csc^{2 }*x*

tan^{2 }*x* + 1 = sec^{2 }*x*

To limit the amount of memorizing you have to do, you can use the first Pythagorean identity to derive the other two:

If you divide every term of sin^{2} *θ* + cos^{2} *θ* = 1 by sin^{2}* θ*, you get

, which simplifies to another Pythagorean identity:

1 + cot^{2}* θ* = csc^{2}* θ*, because

(because of the reciprocal identity)

(also because of the reciprocal identity)

When you divide every term of sin^{2} *θ* + cos^{2} *θ *= 1 by cos^{2}* θ*, you get

, which simplifies to the third Pythagorean identity:

tan^{2 }*θ* + 1 = sec^{2 }*θ*, because

**Putting the Pythagorean identities into action**

You normally use Pythagorean identities if you know one function and are looking for another. For example, if you know the sine ratio, you can use the first Pythagorean identity from the previous section to find the cosine ratio. In fact, you can find whatever you’re asked to find if all you have is the value of one trig function and the understanding of what quadrant the angle *θ* is in.

For example, if you know that sin *θ* = 24/25 and π/2 < *θ *< π, you can find cos *θ* by following these steps:

**1.** **Plug what you know into the appropriate Pythagorean identity.**

Because you’re using sine and cosine, you use the first identity: sin^{2} *θ* + cos^{2} *θ* = 1. Plug in the values you know to get (24/25)^{2} + cos^{2} *θ* = 1.

**2.** **Isolate the trig function with the variable on one side.**

First square the sine value to get 576/625, giving you 576/625 + cos^{2} *θ* = 1. Subtract 576/625 from both sides (Hint: You need to find a common denominator): cos^{2} *θ* = 49/625.

**3.** **Square root both sides to solve.**

You now have cos *θ* = ±7/25. But you can have only one solution because of the constraint π/2 < *θ* < π you’re given in the problem.

**4.** **Draw a picture of the unit circle so you can visualize the angle.**

Because π/2 < *θ* < π, the angle lies in quadrant II, so the cosine of *θ* must be negative. You have your answer: cos *θ* = –7/25.

**Using the Pythagorean identities to prove an equality**

The Pythagorean identities pop up frequently in trig proofs. Pay attention and look for trig functions being squared. Try changing them to a Pythagorean identity and see whether anything interesting happens. This section shows you how one proof can involve a Pythagorean identity.

After you change sines and cosines, the proof simplifies and makes your job

that much easier. For example, follow these steps to prove :

**1.** **Convert all the functions in the equality to sines and cosines.**

**2.** **Use the properties of fractions to simplify.**

Dividing by a fraction is the same as multiplying by its reciprocal, so

sin^{2 }*x* + cos^{2 }*x* = 1

**3.** **Identify the Pythagorean identity on the left side of the equality.**

Because sin^{2 }*x* + cos^{2 }*x* = 1, you can say that 1 = 1.

**Even/odd identities**

Because sine, cosine, and tangent are functions (trig functions), they can be defined as even or odd functions as well (see Chapter 3). Sine and tangent are both odd functions, and cosine is an even function. In other words,

sin(–*x*) = –sin *x*

cos(–*x*) = cos *x*

tan(–*x*) = –tan *x*

These identities will all make appearances in problems that ask you to simplify an expression, prove an identity, or solve an equation (see Chapter 6). The big red flag this time? The fact that the variable inside the trig function is negative. When tan(–*x*), for example, appears somewhere in an expression, it should usually be changed to –tan *x.*

**Simplifying expressions with even/odd identities**

Mostly, you use even/odd identities for graphing purposes, but you may see them in simplifying problems as well. (You can find the graphs of the trigonometric equations in Chapter 7 if you need a refresher.) You use an even/odd identity to simplify any expression where –*x* (or whatever variable you see) is inside the trig function.

In the following list, we show you how to simplify [1 + sin(–*x*)][1 – sin(–*x*)]:

**1.** **Get rid of all the – x values inside the trig functions.**

You see two sin(–*x*) functions, so you replace them both with –sin *x* to get [1 + (–sin *x*)][1 – (–sin *x*)].

**2.** **Simplify the new expression.**

First adjust the two negative signs within the parentheses to get (1 – sin *x*)(1 + sin *x*), and then FOIL these two binomials to get 1 – sin^{2 }*x.*

**3.** **Look for any combination of terms that could give you a Pythagorean identity.**

Whenever you see a function squared, you should think of the Pythagorean identities. Looking back at the section “Pythagorean identities,” you see that 1 – sin^{2 }*x* is the same as cos^{2 }*x.* Now the expression is fully simplified as cos^{2 }*x.*

**Proving an equality with even/odd identities**

When asked to prove an identity, if you see a negative variable inside a trig function, you automatically use an even/odd identity. First, replace all trig functions with –*θ* inside the parentheses. Then simplify the trig expression to make one side look like the other side. Here’s just one example of how this works.

With the following steps, prove this identity:

**1.** **Replace all negative angles and their trig functions with the even/odd identity that matches.**

**2.** **Simplify the new expression.**

Because the right side doesn’t have any fractions in it, eliminating the fractions from the left side is an excellent place to start. In order to subtract fractions, you first must find a common denominator. However, before doing that, look at the first term of the fraction. This fraction can be split up into the sum of two fractions, as can the second fraction. By doing this step first, certain terms simplify and make your job much easier when the time comes to work with the fractions.

Therefore, you get

which quickly simplifies to

Now you must find a common denominator. For this example, the common denominator is sin *θ* · cos *θ*. Multiplying the first term by

and the second term by

gives you

You can rewrite this equation as

Here is a Pythagorean identity in its finest form! Sin^{2 }*θ* + cos^{2} *θ* = 1 is the most frequently used of the Pythagorean identities. This equation then simplifies to

Using the reciprocal identities, you get

So sec *θ *· csc *θ* = sec *θ *· csc *θ*.

**Co-function identities**

If you take the graph of sine and shift it to the left or right, it looks exactly like the cosine graph (see Chapter 7). The same is true for tangent and cotangent, as well as secant and cosecant. That’s the basic premise of *co-function identities* — they say that the sine and cosine functions have the same values, but those values are shifted slightly on the coordinate plane when you look at one function compared to the other. You have experience with all six of the trig functions, as well as their relationships to one another. The only difference is that in this section, we introduce them formally as *identities.*

The following list of co-function identities illustrates this point:

sin *x* = cos(π/2 – *x*)

cos *x* = sin(π/2 – *x*)

tan *x* = cot(π/2 – *x*)

cot *x* = tan(π/2 – *x*)

sec *x* = csc(π/2 – *x*)

csc *x* = sec(π/2 – *x*)

**Putting co-function identities to the test**

The co-function identities are great to use whenever you see π/2 inside the grouping parentheses. You may see functions in the expressions such as sin(π/2 – *x*). If the quantity inside the trig function looks like (π/2 – *x*) or (90° – *θ*), you’ll know to use the co-function identities.

For example, to simplify , follow these steps:

**1.** **Look for co-function identities and substitute.**

First realize that cos(π/2 – *x*) is the same as sin *x* because of the co-function identity. That means you can substitute sin *x* in for cos(π/2 – *x*) to get

**2. Look for other substitutions you can make.**

Because of the reciprocal identity for cotangent, is the same as cot *x.*

**Proving an equality by employing the co-function identities**

Co-function identities also pop up in trig proofs. If you see the expression π/2 – *x* in parentheses inside any trig function, you know to use a co-function identity for the proof. Follow the steps to prove this equality:

**1.** **Replace any trig functions with **π**/2 in them with the appropriate co-function identity.**

Replacing csc(π/2 – *θ*) with sec(*θ*), you get

**2.** **Simplify the new expression.**

You have many trig identities at your disposal, and you may use any of them at any given time. Now is the perfect time to use an even/odd identity for tangent:

Then use the reciprocal identity for secant and the sines and cosines definition for tangent to get

Finally, rewrite this complex fraction as the division of two simpler fractions:

Cancel anything that’s both in the numerator and the denominator and then simplify. This step gives you

Rewrite the last line of the proof as –csc *θ* = –csc *θ*.

**Periodicity identities**

*Periodicity identities* illustrate how shifting the graph of a trig function by one period to the left or right results in the same function. (We cover periods and periodicity identities when we show you how to graph trig functions in Chapter 7.) The functions of sine, cosine, secant, and cosecant repeat every 2π; tangent and cotangent, on the other hand, repeat every π.

The following identities show how the different trig functions repeat:

sin(*x* + 2π) = sin *x*

cos(*x* + 2π) = cos *x*

tan(*x* + π) = tan *x*

cot(*x* + π) = cot *x*

sec(*x* + 2π) = sec *x*

csc(*x* + 2π) = csc *x*

**Seeing how the periodicity identities work to simplify equations**

Similar to the co-function identities, you use the periodicity identities when you see (*x* + 2π) or (*x* – 2π) inside a trig function. Because adding (or subtracting) 2π radians from an angle gives you a new angle in the same position, you can use that idea to form an identity. For tangent and cotangent only, adding or subtracting π radians from the angle gives you the same result, because the period of the tangent and cotangent functions is π.

For example, to simplify sin(2π + *θ*) + cos(2π + *θ*) · cot(π + *θ*), follow these steps:

**1.** **Replace all trig functions with 2**π** inside the parentheses with the appropriate periodicity identity.**

For this example, sin *θ* + cos *θ* · cot *θ*.

**2.** **Simplify the new expression.**

Start with this expression:

To find a common denominator to add the fractions, multiply the first

term by . Here’s the new fraction:

Add them together to get this:

You can see a Pythagorean identity in the numerator, so replace sin^{2} *θ* + cos^{2} *θ* with 1. Therefore, the fraction becomes

**Proving an equality with the periodicity identities**

Using the periodicity identities also comes in handy when you need to prove an equality that includes the expression (*x* + 2π) or the addition (or subtraction) of the period. For example, to prove [sec(2π + *x*) – tan(π + *x*)][csc(2π + *x*) + 1] = cot *x,* follow these steps:

**1.** **Replace all trig functions with the appropriate periodicity identity.**

You’re left with (sec *x* – tan *x*)(csc *x*+ 1).

**2.** **Simplify the new expression.**

For this example, the best place to start is to FOIL (see Chapter 4):

(sec *x* · csc *x* + sec *x* · 1 – tan *x* · csc *x* – tan *x* · 1)

Now convert all terms to sines and cosines to get

Then find a common denominator and add the fractions:

**3.** **Apply any other applicable identities.**

You have a Pythagorean identity in the form of 1 – sin^{2} *x,* so replace it with cos^{2} *x. *Cancel one of the cosines in the numerator (because it’s squared) with the cosine in the denominator to get

Finally, this equation simplifies to cot *x* = cot *x.*

**Tackling Difficult Trig Proofs: Some Techniques to Know**

Historically speaking, most of our students have struggled with (and hated) proofs, so we’ve dedicated this section entirely to them (the proofs *and *the students). So far in this chapter, we’ve shown you proofs that require only a few basic steps to complete. Now we show you how to tackle the more complicated proofs. The techniques here are based on ideas you’ve dealt with before in your math journey. Okay, so a few trig functions are thrown into the discussion, but why should that scare you?

One tip will always help you when you’re faced with complicated trig proofs that require multiple identities: *Always* check your work and review all the identities you know to make sure that you haven’t forgotten to simplify something.

The goal in proofs is to make one side of the given equation look like the other through a series of steps, all of which are based on identities, properties, and definitions. No cheating by making up something that doesn’t exist to get done! All decisions you make must be based on the rules. Here’s an overview of the techniques we show you in this section:

**Fractions in proofs:** These types of proofs come equipped with every rule you’ve ever learned regarding fractions, plus you have to deal with identities on top of that.

**Factoring: **Degrees higher than 1 on a trig function often are great indicators that you need to do some factoring. And if you see multiple sets of parentheses, you may have to do some FOILing.

**Square roots: **When roots show up in a proof, sooner or later you’ll probably have to square both sides to get things moving.

**Working on both sides at once:** Sometimes you may get stuck while working on one side of a proof. At that point, we recommend transitioning to the other side.

**Dealing with dreaded denominators**

Most students hate fractions. We don’t know why; they just do! But in dealing with trig proofs, fractions inevitably pop up. So allow us to throw you into the deep end of the pool. Even if you don’t mind fractions, we suggest you still read this section because it shows you specifically how to work with fractions in trig proofs. Three main types of proofs you’ll work with have fractions:

Proofs where you end up creating fractions

Proofs that start off with fractions

Proofs that require multiplying by a conjugate to deal with a fraction

We break down each one of these types in this section with an example proof so you can see what to do.

**Creating fractions when working with reciprocal identities**

We often like to mention how converting all the functions to sines and cosines makes a trig proof easier. When terms are multiplying, this conversion usually allows you to cancel and simplify to your heart’s content so that one side of the equation ends up looking just like the other, which is the goal. But when the terms are adding or subtracting, you may create fractions where none were before. This is especially true when dealing with secant and cosecant,

because you create fractions when you convert them (respectively) to

and . The same is true for tangent when you change it to , and

cotangent becomes .

Here’s an example that illustrates our point. Follow these steps to prove that sec^{2 }*t* + csc^{2 }*t* = sec^{2 }*t* · csc^{2 }*t:*

**1.** **Convert all the trig functions to sines and cosines.**

On the left side, you now have

**2.** **Find the LCD of the two fractions.**

This multiplication gives you

**3.** **Add the two fractions.**

**4.** **Simplify the expression with a Pythagorean identity in the numerator.**

**5.** **Use reciprocal identities to invert the fraction.**

Both sides now have multiplication:

csc^{2 }*t *· sec^{2 }*t *= sec^{2 }*t *· csc^{2 }*t*

Some pre-calculus teachers let you stop there; others, however, make you rewrite the equation so that the left and right sides match exactly. Every teacher has his or her own way of proving trig identities. Make sure that you adhere to your teacher’s expectations; otherwise, you may lose points on a test.

**6.** **Use the properties of equality to rewrite.**

The commutative property of multiplication (see Chapter 1) says that *a* · *b* = *b* · *a,* so sec^{2 }*t* · csc^{2 }*t* = sec^{2 }*t* · csc^{2 }*t.*

**Starting off with fractions**

When the expression you’re given begins with fractions, most of the time you have to add (or subtract) them to get things to simplify. Here’s one example of a proof where doing just that gets the ball rolling. For example, you have to find the LCD to add the two fractions in order to simplify this expression:

With that as the beginning step, follow along:

**1.** **Find the LCD of the two fractions you must add.**

The least common denominator is (1 + sin *t*) · cos *t,* so multiply the first term by

and multiply the second term by

You get

**2.** **Multiply or distribute in the numerators of the fractions.**

**3.** **Add the two fractions.**

**4.** **Look for any trig identities and substitute.**

You can rewrite the numerator as

which is equal to

because cos^{2} *t* + sin^{2} *t* = 1 (a Pythagorean identity).

**5.** **Cancel or reduce the fraction.**

After the top and the bottom are completely factored (see Chapter 4), you can cancel terms:

**6.** **Change any reciprocal trig functions.**

**Multiplying by a conjugate**

When one side of a proof is a fraction with a binomial in its denominator, always consider multiplying by the conjugate before you do anything else. Most of the time, this technique allows you to simplify. When it doesn’t, you’re left to your own devices (and the other techniques presented in this section) to make the proof work.

For example, follow the steps to rewrite this expression without a fraction:

**1.** **Multiply by the conjugate of the denominator.**

The conjugate of *a* + *b* is *a* – *b,* and vice versa. So you have to multiply by sec *θ* + 1 on the top and bottom of the fraction. This step gives you

**2.** **FOIL the conjugates.**

If you’ve been following along all chapter, the bottom should look awfully familiar. One of those Pythagorean identities? Yep!

**3.** **Change any identities to their simpler form.**

Using the identity on the bottom, you get

**4.** **Change every trig function to sines and cosines.**

Here it gets more complex:

**5.** **Change the big division bar to a division sign, and then invert the fraction so you can multiply instead.**

.

**6.** **Cancel what you can from the expression.**

The sine on the top cancels one of the sines on the bottom, leaving you with the following equation:

**7.** **Distribute and watch what happens!**

Through cancellations, you go from

to

This expression finally simplifies to cot *θ* + cot *θ* · cos *θ*. And if you’re asked to take it even a step further, you can factor to get cot *θ* (1 + cos *θ*).

**Going solo on each side**

Sometimes doing work on both sides of a proof, one side at a time, leads to a quicker solution, because in order to prove a very complicated identity, you may need to complicate the expression even further before it can begin to simplify. However, you should take this action only in dire circumstances after every other technique has failed (but don’t tell your teacher we told you so!).

The main idea is that you work on the left side first, stop when you just can’t go any further, and then switch to working on the right side. By switching back and forth, your goal is to make the two sides of the proof meet in the middle somewhere.

For example, follow the steps to prove this identity:

**1.** **Break up the fraction by writing each term in the numerator over the term in the denominator, separately.**

The rules of fractions state that when only one term sits in the denominator, you can do this step because each part on top is being divided by the bottom.

You now have

**2.** **Use reciprocal rules to simplify.**

The first fraction on the left side is the reciprocal of tan *θ*, and cot *θ* divided by itself is 1. You now have

You’ve come to the end of the road on the left side. The expression is now so simplified that it would be hard to expand it again to look like the right side, so you should turn to the right side and simplify it.

**3.** **Look for any applicable trig identities on the right side.**

You use a Pythagorean identity to identify that csc^{2}* θ* = 1 + cot^{2}* θ*.

You now have tan *θ *+ 1 = tan *θ *+ 1 + cot^{2}* θ* – cot^{2} *θ*.

**4.** **Cancel where possible.**

Aha! The right side has cot^{2 }*θ* – cot^{2 }*θ*, which is 0! Cancel them to leave only tan *θ *+ 1 = tan *θ *+ 1.

**5. Rewrite the proof starting on one side and ending up like the other side.**

Some pre-calc teachers do not accept working on both sides of an equation as a valid proof. If you’re unfortunate enough to encounter a teacher like this, we suggest that you still work on both sides of the equation, but for your eyes only. Be sure to rewrite your work for your teacher by simply going down one side and up the other (like we did in Step 5).