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# 3.2: Proofs of Trigonometric Identities

Difficulty Level: At Grade Created by: CK-12
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Practice Proofs of Trigonometric Identities

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What if your instructor gave you two trigonometric expressions and asked you to prove that they were true. Could you do this? For example, can you show that

sin2θ=1cos2θ2\begin{align*}\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\end{align*}

### Trigonometric Identities

In Trigonometry you will see complex trigonometric expressions. Often, complex trigonometric expressions can be equivalent to less complex expressions. The process for showing two trigonometric expressions to be equivalent (regardless of the value of the angle) is known as validating or proving trigonometric identities.

There are several options a student can use when proving a trigonometric identity.

Option One: Often one of the steps for proving identities is to change each term into their sine and cosine equivalents.

Option Two: Use the Trigonometric Pythagorean Theorem and other Fundamental Identities.

Option Three: When working with identities where there are fractions- combine using algebraic techniques for adding expressions with unlike denominators.

Option Four: If possible, factor trigonometric expressions. For example, 2+2cosθsinθ(1+cosθ)=2cscθ\begin{align*}\frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} = 2 \csc \theta\end{align*} can be factored to 2(1+cosθ)sinθ(1+cosθ)=2cscθ\begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*} and in this situation, the factors cancel each other.

#### Proving Identities

1. Prove the identity: cscθ×tanθ=secθ\begin{align*}\csc \theta \times \tan \theta = \sec \theta\end{align*}

Reducing each side separately. It might be helpful to put a line down, through the equals sign. Because we are proving this identity, we don’t know if the two sides are equal, so wait until the end to include the equality.

cscx×tanx1sinx×sinxcosx1sinx×sinxcosx1cosxsecx1cosx1cosx1cosx\begin{align*}\begin{array}{c|c } \csc x \times \tan x & \sec x \\ \frac{1}{\sin x} \times \frac{\sin x}{\cos x} & \frac{1}{\cos x} \\ \frac{1}{\cancel{\sin x}} \times \frac{\cancel{\sin x}}{\cos x}& \frac{1}{\cos x} \\ \frac{1}{\cos x} & \frac{1}{\cos x} \end{array}\end{align*}

At the end we ended up with the same thing, so we know that this is a valid identity.

Notice when working with identities, unlike equations, conversions and mathematical operations are performed only on one side of the identity. In more complex identities sometimes both sides of the identity are simplified or expanded. The thought process for establishing identities is to view each side of the identity separately, and at the end to show that both sides do in fact transform into identical mathematical statements.

2. Prove the identity: (1cos2x)(1+cot2x)=1\begin{align*}(1-\cos^2 x)(1+\cot^2 x) = 1\end{align*}

Use the Pythagorean Identity and its alternate form. Manipulate sin2θ+cos2θ=1\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*} to be sin2θ=1cos2θ\begin{align*}\sin^2 \theta = 1 - \cos^2 \theta\end{align*}. Also substitute csc2x\begin{align*}\csc^2 x\end{align*} for 1+cot2x\begin{align*}1+ \cot^2 x\end{align*}, then cross-cancel.

(1cos2x)(1+cot2x)sin2xcsc2xsin2x1sin2x11111\begin{align*}\begin{array}{c|c } (1-\cos^2 x)(1+\cot^2 x) & 1 \\ \sin^2 x \cdot \csc^2 x & 1 \\ \sin^2 x \cdot \frac{1}{\sin^2 x}& 1 \\ 1 & 1 \end{array}\end{align*}

3. Prove the identity: sinθ1+cosθ+1+cosθsinθ=2cscθ\begin{align*}\frac{\sin \theta}{1+ \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta\end{align*}.

Combine the two fractions on the left side of the equation by finding the common denominator: (1+cosθ)×sinθ\begin{align*}(1 + \cos \theta) \times \sin \theta\end{align*}, and the change the right side into terms of sine.

sinθ1+cosθ+1+cosθsinθsinθsinθsinθ1+cosθ  +1+cosθsinθ1+cosθ1+cosθ sin2θ+(1+cosθ)2sinθ(1+cosθ)2cscθ2cscθ2cscθ\begin{align*}\begin{array}{c|c } \frac{\sin \theta}{1+ \cos \theta} + \frac{1+ \cos \theta}{\sin \theta} & 2 \csc \theta \\ \frac{\sin \theta}{\sin \theta} \cdot \frac{\sin \theta}{1+ \cos \theta}\ \ + \frac{1+ \cos \theta}{\sin \theta} \cdot \frac{1+ \cos \theta}{1+ \cos \theta} & 2 \csc \theta \\ \quad \ \frac{\sin^2 \theta + (1+ \cos \theta)^2}{\sin \theta (1 + \cos \theta)}& 2 \csc \theta \end{array}\end{align*}

Now, we need to apply another algebraic technique, FOIL. (FOIL is a memory device that describes the process for multiplying two binomials, meaning multiplying the First two terms, the Outer two terms, the Inner two terms, and then the Last two terms, and then summing the four products.) Always leave the denominator factored, because you might be able to cancel something out at the end.

sin2θ+1+2cosθ+cos2θsinθ(1+cosθ)2cscθ\begin{align*}\begin{array}{c|c } \frac{\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta}{\sin \theta (1+ \cos \theta)}& 2 \csc \theta\end{array}\end{align*}

Using the second option, substitute sin2θ+cos2θ=1\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*} and simplify.

1+1+2cosθsinθ(1+cosθ)2+2cosθsinθ(1+cosθ)2(1+cosθ)sinθ(1+cosθ)2sinθ2cscθ2cscθ2cscθ2sinθ\begin{align*}\begin{array}{c|c } \frac{1 + 1 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} & 2 \csc \theta \\ \frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} & 2 \csc \theta \\ \frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)}& 2 \csc \theta \\ \frac{2}{\sin \theta}& \frac{2}{\sin \theta} \end{array}\end{align*}

Option Four: If possible, factor trigonometric expressions. Actually procedure four was used in #2: 2+2cosθsinθ(1+cosθ)=2cscθ\begin{align*}\frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} = 2 \csc \theta\end{align*} can be factored to 2(1+cosθ)sinθ(1+cosθ)=2cscθ\begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*} and in this situation, the factors cancel each other.

### Examples

#### Example 1

Earlier, you were asked to prove

sin2θ=1cos2θ2\begin{align*}\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\end{align*}

First remember the Pythagorean Identity:

sin2θ+cos2θ=1\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*}

Therefore,

sin2θ=1cos2θ\begin{align*}\sin^2 \theta = 1 - \cos^2 \theta\end{align*}

From the Double Angle Identities, we know that

cos2θ=cos2θsin2θcos2θ=cos2θ+sin2θ\begin{align*} \cos 2\theta = \cos^2 \theta - \sin^2 \theta\\ \cos^2 \theta = \cos 2\theta + \sin^2 \theta\\ \end{align*}

Substituting this into the above equation for sin2\begin{align*}\sin^2\end{align*},

sin2θ=1(cos2θ+sin2θ)sin2θ=1cos2θsin2θ2sin2θ=1cos2θsin2θ=1cos2θ2\begin{align*} \sin^2 \theta = 1 - (\cos 2\theta + \sin^2 \theta)\\ \sin^2 \theta = 1 - \cos 2\theta - \sin^2 \theta\\ 2\sin^2 \theta = 1 - \cos 2\theta\\ \sin^2 \theta = \frac{1 - \cos 2\theta}{2}\\ \end{align*}

#### Example 2

Prove the identity: sinxtanx+cosx=secx\begin{align*}\sin x \tan x + \cos x = \sec x\end{align*}

Step 1: Change everything into sine and cosine

sinxtanx+cosxsinxsinxcosx+cosx=secx=1cosx\begin{align*}\sin x \tan x + \cos x & = \sec x \\ \sin x \cdot \frac{\sin x}{\cos x} + \cos x & = \frac{1}{\cos x} \end{align*}

Step 2: Give everything a common denominator, cosx\begin{align*}\cos x\end{align*}.

sin2xcosx+cos2xcosx=1cosx\begin{align*}\frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{1}{\cos x}\end{align*}

Step 3: Because the denominators are all the same, we can eliminate them.

sin2x+cos2x=1\begin{align*}\sin^2x + \cos^2x = 1\end{align*}

We know this is true because it is the Trig Pythagorean Theorem

#### Example 3

Prove the identity: cosxcosxsin2x=cos3x\begin{align*}\cos x - \cos x \sin^2 x = \cos^3 x\end{align*}

Step 1: Pull out a cosx\begin{align*}\cos x\end{align*}

cosxcosxsin2xcosx(1sin2x)=cos3x=cos3x\begin{align*}\cos x - \cos x \sin^2x & = \cos^3x \\ \cos x (1 - \sin^2x) & = \cos^3 x\end{align*}

Step 2: We know sin2x+cos2x=1\begin{align*}\sin^2x + \cos^2x = 1\end{align*}, so cos2x=1sin2x\begin{align*}\cos^2x = 1 - \sin^2x\end{align*} is also true, therefore cosx(cos2x)=cos3x\begin{align*}\cos x (\cos^2x) = \cos^3x\end{align*}. This, of course, is true, we are finished!

#### Example 4

Prove the identity: sinx1+cosx+1+cosxsinx=2cscx\begin{align*}\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x\end{align*}

Step 1: Change everything in to sine and cosine and find a common denominator for left hand side.

sinx1+cosx+1+cosxsinx=2cscxsinx1+cosx+1+cosxsinx=2sinxLCD: sinx(1+cosx)sin2x+(1+cosx)2sinx(1+cosx)\begin{align*}& \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x \\ & \frac{\sin x}{1+\cos x} + \frac{1 + \cos x}{\sin x} = \frac{2}{\sin x} \leftarrow \text{LCD}:\ \sin x (1 + \cos x) \\ & \frac{\sin^2 x + (1 + \cos x)^2}{\sin x (1 + \cos x)}\end{align*}

Step 2: Working with the left side, FOIL and simplify.

sin2x+1+2cosx+cos2xsinx(1+cosx)sin2x+cos2x+1+2cosxsinx(1+cosx)1+1+2cosxsinx(1+cosx)2+2cosxsinx(1+cosx)2(1+cosx)sinx(1+cosx)2sinxFOIL (1+cosx)2move cos2xsin2x+cos2x=1addfactor out 2cancel (1+cosx)\begin{align*}& \frac{\sin^2 x + 1 + 2 \cos x + \cos^2 x}{\sin x (1 + \cos x)} && \rightarrow \text{FOIL}\ (1 + \cos x)^2 \\ & \frac{\sin^2 x + \cos^2 x + 1 + 2 \cos x }{\sin x (1 + \cos x)} && \rightarrow \text{move}\ \cos^2 x \\ & \frac{1 + 1 + 2 \cos x}{\sin x (1 + \cos x)} && \rightarrow \sin^2 x + \cos^2 x = 1 \\ & \frac{2 + 2 \cos x}{\sin x (1 + \cos x)} && \rightarrow \text{add} \\ & \frac{2(1+ \cos x)}{\sin x (1 + \cos x)} && \rightarrow \text{factor out}\ 2 \\ & \frac{2}{\sin x} && \rightarrow \text{cancel}\ (1 + \cos x)\end{align*}

### Review

Use trigonometric identities to simplify each expression as much as possible.

1. tan(x)cos(x)\begin{align*}\tan(x)\cos(x)\end{align*}
2. cos(x)cos3(x)\begin{align*}\cos(x)-\cos^3(x)\end{align*}
3. 1cos2(x)sin(x)\begin{align*}\frac{1-\cos^2(x)}{\sin(x)}\end{align*}
4. cot(x)sin(x)\begin{align*}\cot(x)\sin(x)\end{align*}
5. 1sin2(x)cos(x)\begin{align*}\frac{1-\sin^2(x)}{\cos(x)}\end{align*}
6. sin(x)csc(x)\begin{align*}\sin(x)\csc(x)\end{align*}
7. tan(x)cot(x)\begin{align*}\tan(-x)\cot(x)\end{align*}
8. sec2(x)tan2(x)cos2(x)+sin2(x)\begin{align*}\frac{\sec^2(x)-\tan^2(x)}{\cos^2(x)+\sin^2(x)}\end{align*}

Prove each identity.

1. tan(x)+cot(x)=sec(x)csc(x)\begin{align*}\tan(x)+\cot(x)=\sec(x)\csc(x)\end{align*}
2. sin(x)=sin2(x)+cos2(x)csc(x)\begin{align*}\sin(x)=\frac{\sin^2(x)+\cos^2(x)}{\csc(x)}\end{align*}
3. 1sec(x)1+1sec(x)+1=2cot(x)csc(x)\begin{align*}\frac{1}{\sec(x)-1}+\frac{1}{\sec(x)+1}=2\cot(x)\csc(x)\end{align*}
4. \begin{align*}(\cos(x))(\tan(x)+\sin(x)\cot(x))=\sin(x)+\cos^2(x)\end{align*}
5. \begin{align*}\sin^4(x)-\cos^4(x)=\sin^2(x)-\cos^2(x)\end{align*}
6. \begin{align*}\sin^2(x)\cos^3(x)=(\sin^2(x)-\sin^4(x))(\cos(x))\end{align*}
7. \begin{align*}\frac{\sin(x)}{\csc(x)}=1-\frac{\cos(x)}{\sec(x)}\end{align*}

To see the Review answers, open this PDF file and look for section 3.2.

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Color Highlighted Text Notes

### Vocabulary Language: English

FOIL

FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms.

Trigonometric Identity

A trigonometric identity is an equation that relates two or more trigonometric functions.

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