---
title: "Precalc: Verifying Trig Identities"
---

## Trig Identities Cheatsheet

First, let's remember our **negative angle rules** (**N** for **N**egative), since it turns out you can always use these rules to *derive* any of the identities below from any of the other ones!

(**N1**: Negative Angle Negates $\sin$)

$$
\sin(-\theta) = -\sin(\theta)
$$

(**N2**: Negative Angle Doesn't Change $\cos$)

$$
\cos(-\theta) = \cos(\theta)
$$

The way I split these is into three groups of identities:

### 1. Addition/Subtraction Identities (**A** for **A**ddition)

(**A1**: Addition of Sines)

$$
\sin( \alpha \pm \beta ) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)
$$

(**A2**: Addition of Cosines)

$$
\cos( \alpha + \beta ) = \cos(\alpha)\sin(\beta) - \sin(\alpha)\cos(\beta)
$$

(**A3**: Addition of Tangents)

$$
\tan( \alpha + \beta ) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}, \; \; \tan( \alpha - \beta ) = \frac{\tan(\alpha) - \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}
$$

### 2. Multiplication/Division Identities (**M** for **M**ultiplicaton)

(**M1**: Sine Double Angle)

$$
\sin( 2 \theta ) = 2 \sin( \theta ) \cos( \theta )
$$

(**M2**: Cosine Double Angle)

$$
\cos( 2 \theta ) = \cos^2( \theta ) - \sin^2( \theta ) \\
\phantom{\cos} = 1 - 2\sin^2(\theta) \\
\phantom{\cos} = 2\cos^2(\theta) - 1
$$

(**M3**: Tangent Double Angle)

$$
\tan(2 \theta) = \frac{ 2 \tan( \theta) }{ 1-\tan(\theta) }
$$

### 3. Squared Identities (**S** for **Squared**)

(Also known as the **Pythagorean Identities**)

(**S1**: Pythagorean Identity)

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

(**S2**: Pythagorean Tangent-Squared)

$$
1 + \tan^2(\theta) = \sec^2(\theta)
$$

(**S3**: Pythagorean Cotangent-Squared)

$$
\cot^2(\theta) + 1 = \csc^2(\theta)
$$

## Practice Problems

### Problem 1: Summary from Today's Session

1. From today: Verify the identity you have, that

$$
\cos( 2 \theta ) = 2 \cos^2(\theta) - 1
$$

Though we saw today how there are many, many ways you could verify this, using the identities you know, we also saw that probably the fastest way would be using the following steps (to go from the left-hand side $\cos(2\theta)$ to the right-hand side $2\cos^2(\theta) - 1$):

a. Break $\cos(2\theta)$ into $\cos(\theta + \theta)$, since this transforms the *multiplication* happening inside our $\cos$ into an *addition*, meaning that we can use our Addition Identities from above.

$$
\cos( 2 \theta ) = \cos( \theta + \theta )
$$

b. Use our identity (**A2**: Addition of Cosines) to turn this addition *inside* the cosine into an addition of trig functions *without* any additions happening inside of them. With $\theta$ as both our $\alpha$ and $\beta$ values, we get

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

c. We're close, since we now have the $\cos^2(\theta)$ term that we wanted, but it came with an extra $\sin^2(\theta)$ term that we need to get rid of. So, since we see that the **Pythagorean Identities** are the ones that can help us when we have squared terms, let's take a look at the main one:

$$
\cos^2(\theta) + \sin^2(\theta) = 1.
$$

d. We know we want to get rid of the $\sin^2(\theta)$ term in $\cos^2(\theta) - \sin^2(\theta)$, so let's solve this Pythagorean identity for $\sin^2(\theta)$, which will give us a way to rewrite this term:

$$
\cos^2(\theta) + \sin^2(\theta) = 1 \implies \sin^2(\theta) = 1 - \cos^2(\theta).
$$

So, we've derived a **new** identity, that anywhere we see $\sin^2(\theta)$ we can always rewrite it as $1 - \cos^2(\theta)$! So, let's rewrite the $\sin^2(\theta)$ term in the expression we had arrived at above:

$$
\cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - \left(1 - \cos^2(\theta)\right) = \cos^2(\theta) - 1 + \cos^2(\theta) = 2\cos^2(\theta) - 1,
$$

and we're done! We started from $\cos(2\theta)$, and used valid identities (and stuff derived from these identities) throughout to arrive at $2\cos^2(\theta) - 1$, as desired. :heavy_check_mark:

### Problem 2: Lots of cancelling and simplifying

2. The one you mentioned from class: Verify that

$$
\frac{\sin(\theta) - 1}{\cos(\theta)} = \frac{\cos(\theta) - \cot(\theta)}{\cos(\theta) \cdot \cot(\theta)}.
$$

In this case, as you mentioned in how you solved it, it isn't so bad identity-wise, instead it really requires a lot of algebra and cancelling stuff out. Starting from the right-hand side this time, we carry out the following steps (where I'm not including full explanations just because, you've already solved it! But, wanted to include so we have a bunch of examples in one place)

a. Transform $\cot(\theta)$ into $\frac{\cos(\theta)}{\sin(\theta)}$:

$$
\frac{\cos(\theta) - \cot(\theta)}{\cos(\theta) \cdot \cot(\theta)} = \frac{\cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)}}{\cos(\theta) \cdot \frac{\cos(\theta)}{\sin(\theta)}}.
$$

b. Combine the $\cos(\theta)$ terms in the denominator, so that it becomes a single fraction:

$$
\frac{\cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)}}{\cos(\theta) \cdot \frac{\cos(\theta)}{\sin(\theta)}} = \frac{\cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)}}{\frac{\cos^2(\theta)}{\sin(\theta)}}
$$

c. Now we can "flip" the denominator, since division by a fraction is equivalent to multiplication by its reciprocal:

$$
\frac{\cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)}}{\frac{\cos^2(\theta)}{\sin(\theta)}} = \left( \cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)} \right) \cdot \frac{\sin(\theta)}{\cos^2(\theta)},
$$

d. And now we can multiply the big parentheses term through by $\sin(\theta)$ to get

$$
\frac{\cos(\theta)\sin(\theta) - \cos(\theta)}{\cos^2(\theta)},
$$

e. We can factor a $\cos(\theta)$ out of the numerator to get

$$
\frac{ \cos(\theta) \left( \sin(\theta) - 1 \right) }{ \cos^2(\theta) },
$$

f. And now the cosine in the numerator and *one* of the cosines in the denominator cancel out, giving us

$$
\frac{ \sin(\theta) - 1 }{ \cos(\theta) },
$$

as we wanted to show! :heavy_check_mark:

### Problem 3: Two Different Pathways to the Answer

3. New problem! Verify the identity

$$
\frac{\sec^2(\theta) - 1}{\sec^2(\theta)} = \sin^2(\theta)
$$

Here, like I promised at the end of our session, we have an example where there are two totally-valid paths to the solution!

**Path 1**:

a. Use fraction rules to turn the left-hand side into a subtraction of two fractions:

$$
\frac{\sec^2(\theta) - 1}{\sec^2(\theta)} = \frac{\sec^2(\theta)}{\sec^2(\theta)} - \frac{1}{\sec^2(\theta)}
$$

b. We see that the left fraction (before the minus sign) just turns into a $1$, while the right fraction (after the minus sign) is exactly our definition of $\cos^2(\theta)$ (since $\sec( \theta )$ is *defined as* the reciprocal of $\cos(\theta)$), giving us

$$
\frac{\sec^2(\theta)}{\sec^2(\theta)} - \frac{1}{\sec^2(\theta)} = 1 - \cos^2(\theta).
$$

Now, using the same **Pythagorean identity** reasoning we used in steps c and d of Problem 1 above, we know that we can rewrite $1 - \cos^2(\theta)$ as just $\sin^2(\theta)$, giving us the result we wanted to verify! :heavy_check_mark:

**Path 2**:

Here, at the *beginning*, we could use our (**S3**: Pythogrean Tangent Squared) identity from above, to turn that $\sec^2(\theta)$ into $1 + \tan^2(\theta)$:

$$
\frac{\sec^2(\theta) - 1}{\sec^2(\theta)} = \frac{(1 + \tan^2(\theta)) - 1}{\sec^2(\theta)}.
$$

Note that we only apply this identity to the *numerator*, rather than both numerator and denominator, because the numerator is the part of the problem that has that pesky $-1$ term we're trying to get rid of, to simplify the expression. Since we did that, we see that the $1$ terms in the numerator indeed cancel, giving us

$$
\frac{(1 + \tan^2(\theta)) - 1}{\sec^2(\theta)} = \frac{ \tan^2(\theta) }{ \sec^2(\theta) }.
$$

Now we can just use our definitions of these two functions to simplify our whole expression, write them just in terms of our two basic trig functions $\sin$ and $\cos$:

$$
\frac{ \tan^2(\theta) }{ \sec^2(\theta) } = \frac{\frac{\sin^2(\theta)}{\cos^2(\theta)}}{\frac{1}{\cos^2(\theta)}},
$$

and since dividing by a fraction is equivalent to multiplying by its reciprocal,

$$
\frac{\frac{\sin^2(\theta)}{\cos^2(\theta)}}{\frac{1}{\cos^2(\theta)}} = \frac{\sin^2(\theta)}{\cos^2(\theta)}\cdot \frac{\cos^2(\theta)}{1},
$$

and the $\cos^2(\theta)$ terms cancel, giving us just $\sin^2(\theta)$, as we wanted to verify! :heavy_check_mark:

### Problem 4: Slightly harder problem requiring "conjugate multiplication"

4. Verify

$$
\frac{\cos(\theta)}{ 1 + \sin(\theta) } = \frac{1 - \sin(\theta)}{\cos(\theta)}
$$

a. Remember that "conjugate multiplication" is just that weird trick with fractions where you can sometimes get a *simpler* term by multiplying the numerator and the denominator by the "conjugate" of the denominator: here, let's look at the term on the left-hand side and see what happens if we multiply *both* the numerator and the denominator by the conjugate of the denominator, which is $1 - \sin(\theta)$:

$$
\frac{\cos(\theta)}{ 1 + \sin(\theta) } = \frac{\cos(\theta)}{ 1 + \sin(\theta) } \cdot \left( \frac{1 - \sin(\theta)}{1 - \sin(\theta)} \right) = \frac{\cos(\theta) - \cos(\theta)\sin(\theta)}{1 - \sin^2(\theta)}
$$

b. Even though it seemed like multiplying the denominator by its conjugate might give some really big gross messy term, in fact once we FOILed we found that a bunch of terms cancel, since

$$
(1 + \sin(\theta))(1 - \sin(\theta)) = 1 + \sin(\theta) - \sin(\theta) - \sin^2(\theta),
$$

and the middle two terms cancel each other out.

c. So, with the fraction we're left with, first note that by the same reasoning as in steps c and d of Problem 1 above, we can use our **Pythagorean identity** to rewrite the $1 - \sin^2(\theta)$ in the denominator as $\cos^2(\theta)$:

$$
\frac{\cos(\theta) - \cos(\theta)\sin(\theta)}{1 - \sin^2(\theta)} = \frac{\cos(\theta) - \cos(\theta)\sin(\theta)}{\cos^2(\theta)}.
$$

d. Now, we can also factor a $\cos(\theta)$ term out of the numerator, to get

$$
\frac{\cos(\theta) - \cos(\theta)\sin(\theta)}{\cos^2(\theta)} = \frac{\cos(\theta) \left( 1 - \sin(\theta) \right)}{\cos^2(\theta)}.
$$

e. This means that the $\cos(\theta)$ term in the numerator cancels out *one of* the $\cos(\theta)$ terms in the denominator, so we're left with

$$
\frac{\cos(\theta) \left( 1 - \sin(\theta) \right)}{\cos^2(\theta)} = \frac{ 1 - \sin(\theta) }{ \cos^2(\theta) },
$$

which is what we wanted to show! :heavy_check_mark:

### Additional Problems

Working these out on your own will be so so helpful, if you have additional time before the test!

5. Verify

$$
\frac{ \cot(\theta) }{ \csc(\theta) } = \cos(\theta)
$$

6. Verify

$$
\frac{\sin^2(\theta) − 1}{\tan(\theta)\cdot \sin(\theta) − \tan(\theta)} = \frac{\sin(\theta) + 1}{\tan(\theta)}
$$

7. Verify

$$
\left( 1 − \cos^2(\theta) \right) \left( 1 + \cot^2(\theta) \right) = 1
$$