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# 3.2: Proving Identities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Prove identities using several techniques.

## Working with Trigonometric Identities

During the course, 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:

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

Solution: 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.

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

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

Solution: 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*}

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

Example 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*}.

Solution: 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 the above 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.

Example 4: Prove the identity: 1+tanθ(1+cotθ)=tanθ\begin{align*}\frac{1+ \tan \theta}{(1 + \cot \theta)} = \tan \theta\end{align*}.

Solution: Change cotθ\begin{align*}\cot \theta\end{align*} to 1tanθ\begin{align*}\frac{1}{\tan \theta}\end{align*} and find a common denominator.

1+tanθ(1+1tanθ)1+tanθ(tanθtanθ+1tanθ)=tanθ=tanθor1+tanθtanθ+1tanθ=tanθ\begin{align*}\frac{1+ \tan \theta}{\left (1+ \frac{1}{\tan \theta} \right )} & = \tan \theta \\ \frac{1+ \tan \theta}{\left (\frac{\tan \theta}{\tan \theta} + \frac{1}{\tan \theta} \right )} & = \tan \theta \qquad \text{or} \qquad \frac{1 + \tan \theta}{\frac{\tan \theta + 1}{\tan \theta}} = \tan \theta\end{align*}

Now invert the denominator and multiply.

\begin{align*}\frac{\tan \theta (1 + \tan \theta)}{\tan \theta + 1} & = \tan \theta \\ \tan \theta & = \tan \theta\end{align*}

## Technology Note

A graphing calculator can help provide the correctness of an identity. For example looking at: \begin{align*}\csc x \times \tan x = \sec x\end{align*}, first graph \begin{align*}y = \csc x \times \tan x\end{align*}, and then graph \begin{align*}y = \sec x\end{align*}. Examining the viewing screen for each demonstrates that the results produce the same graph.

To summarize, when verifying a trigonometric identity, use the following tips:

1. Work on one side of the identity- usually the more complicated looking side.
2. Try rewriting all given expressions in terms of sine and cosine.
3. If there are fractions involved, combine them.
4. After combining fractions, if the resulting fraction can be reduced, reduce it.
5. The goal is to make one side look exactly like the other—so as you change one side of the identity, look at the other side for a potential hint to what to do next. If you are stumped, work with the other side. Don’t limit yourself to working only on the left side, a problem might require you to work on the right.

## Points to Consider

• Are there other techniques that you could use to prove identities?
• What else, besides what is listed in this section, do you think would be useful in proving identities?

## Review Questions

Prove the following identities true:

1. \begin{align*}\sin x \tan x + \cos x = \sec x\end{align*}
2. \begin{align*}\cos x - \cos x \sin^2 x = \cos^3 x\end{align*}
3. \begin{align*}\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x\end{align*}
4. \begin{align*}\frac{\sin x}{1+ \cos x} = \frac{1 - \cos x}{\sin x}\end{align*}
5. \begin{align*}\frac{1}{1+ \cos a} + \frac{1}{1 - \cos a} = 2 + 2 \cot^2 a\end{align*}
6. \begin{align*}\cos^4 b - \sin^4 b = 1 - 2 \sin^2 b\end{align*}
7. \begin{align*}\frac{\sin y + \cos y}{\sin y} - \frac{\cos y - \sin y}{\cos y} = \sec y \csc y\end{align*}
8. \begin{align*}(\sec x - \tan x)^2 = \frac{1-\sin x}{1+\sin x}\end{align*}
9. Show that \begin{align*}2 \sin x \cos x = \sin 2x\end{align*} is true using \begin{align*}\frac{5 \pi}{6}\end{align*}.
10. Use the trig identities to prove \begin{align*}\sec x \cot x = \csc x\end{align*}

1. Step 1: Change everything into sine and cosine \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, \begin{align*}\cos x\end{align*}. \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. \begin{align*}\sin^2x + \cos^2x = 1\end{align*} We know this is true because it is the Trig Pythagorean Theorem
2. Step 1: Pull out a \begin{align*}\cos x\end{align*} \begin{align*}\cos x - \cos x \sin^2x & = \cos^3x \\ \cos x (1 - \sin^2x) & = \cos^3 x\end{align*} Step 2: We know \begin{align*}\sin^2x + \cos^2x = 1\end{align*}, so \begin{align*}\cos^2x = 1 - \sin^2x\end{align*} is also true, therefore \begin{align*}\cos x (\cos^2x) = \cos^3x\end{align*}. This, of course, is true, we are done!
3. Step 1: Change everything in to sine and cosine and find a common denominator for left hand side. \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. \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{fator out}\ 2 \\ & \frac{2}{\sin x} && \rightarrow \text{cancel}\ (1 + \cos x)\end{align*}
4. Step 1: Cross-multiply \begin{align*}\frac{\sin x}{1 + \cos x} & = \frac{1 - \cos x}{\sin x} \\ \sin^2 x & = (1 + \cos x)(1 - \cos x)\end{align*} Step 2: Factor and simplify \begin{align*}& \sin^2 x = 1-\cos^2 x \\ & \sin^2 x + \cos^2 x = 1\end{align*}
5. Step 1: Work with left hand side, find common denominator, FOIL and simplify, using \begin{align*}\sin^2x + \cos^2x = 1\end{align*}. \begin{align*}& \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} = 2 +2 \cot^2 x \\ & \qquad \qquad \frac{1 - \cos x + 1 + \cos x}{(1 + \cos x)(1 - \cos x)} \\ & \qquad \qquad \qquad \frac{2}{1 - \cos^2 x} \\ & \qquad \qquad \qquad \quad \frac{2}{\sin^2 x}\end{align*} Step 2: Work with the right hand side, to hopefully end up with \begin{align*}\frac{2}{\sin^2 x}\end{align*}. \begin{align*}& = 2 + 2 \cot^2 x && \\ & = 2 + 2 \ \frac{\cos^2 x}{\sin^2 x} && \\ & = 2 \left (1 + \frac{\cos^2 x}{\sin^2 x} \right ) && \rightarrow \text{factor out the}\ 2 \\ & = 2 \left (\frac{\sin^2 x + \cos^2 x}{\sin^2 x} \right ) && \rightarrow \text{common denominator} \\ & = 2 \left (\frac{1}{\sin^2 x} \right ) && \rightarrow \text{trig pythagorean theorem} \\ & = \frac{2}{\sin^2 x} && \rightarrow \text{simply/multiply}\end{align*} Both sides match up, the identity is true.
6. Step 1: Factor left hand side \begin{align*}\begin{array}{c|c } \qquad \qquad \qquad \ \cos^4 b - \sin^4 b & 1 - 2 \sin^2 b \\ (\cos^2 b + \sin^2 b)(\cos^2 b - \sin^2 b) & 1 - 2 \sin^2 b \\ \qquad \qquad \qquad \ \cos^2 b - \sin^2 b& 1 - 2 \sin^2 b \end{array}\end{align*} Step 2: Substitute \begin{align*}1 - \sin^2 b\end{align*} for \begin{align*}\cos^2b\end{align*} because \begin{align*}\sin^2x + \cos^2x = 1\end{align*}. \begin{align*}\begin{array}{c|c } (1 - \sin^2 b) - \sin^2 b & 1 - 2 \sin^2 b \\ \quad 1 - \sin^2 b - \sin^2 b & 1 - 2 \sin^2 b \\ \qquad \quad 1 - 2\ \ \sin^2 b & 1 - 2 \sin^2 b \end{array}\end{align*}
7. Step 1: Find a common denominator for the left hand side and change right side in terms of sine and cosine. \begin{align*}\frac{\sin y + \cos y}{\sin y} - \frac{\cos y - \sin y}{\cos y} & = \sec y \csc y \\ \frac{\cos y (\sin y + \cos y) - \sin y (\cos y - \sin y)}{\sin y \cos y} = \frac{1}{\sin y \cos y}\end{align*} Step 2: Work with left side, simplify and distribute. \begin{align*}& \frac{\sin y \cos y + \cos^2 y - \sin y \cos y + \sin^2 y}{\sin y \cos y} \\ & \qquad \qquad \quad \ \frac{\cos^2 y + \sin^2 y}{\sin y \cos y} \\ & \qquad \qquad \quad \ \frac{1}{\sin y \cos y}\end{align*}
8. Step 1: Work with left side, change everything into terms of sine and cosine. \begin{align*}& (\sec x - \tan x)^2 = \frac{1 - \sin x}{1 + \sin x} \\ & \left (\frac{1}{\cos x} - \frac{\sin x}{\cos x} \right )^2 \\ & \qquad \left (\frac{1 - \sin x}{\cos x} \right )^2 \\ & \qquad \frac{(1 - \sin x)^2}{\cos^2 x}\end{align*} Step 2: Substitute \begin{align*}1 - \sin^2x\end{align*} for \begin{align*}\cos^2x\end{align*} because \begin{align*}\sin^2x + \cos^2x = 1\end{align*} \begin{align*}\frac{(1 - \sin x)^2}{1 - \sin^2 x} \rightarrow \text{be careful, these are NOT the same!}\end{align*} Step 3: Factor the denominator and cancel out like terms. \begin{align*}& \frac{(1 - \sin x)^2}{(1 + \sin x)(1 - \sin x)} \\ & \qquad \ \frac{1 - \sin x}{1 + \sin x}\end{align*}
9. Plug in \begin{align*}\frac{5 \pi}{6}\end{align*} for \begin{align*}x\end{align*} into the formula and simplify. \begin{align*}2 \sin x \cos x & = \sin 2x \\ 2 \sin \frac{5 \pi}{6} \cos \frac{5 \pi}{6} & = \sin 2 \cdot \frac{5 \pi}{6} \\ 2 \left (\frac{\sqrt{3}}{2} \right ) \left (- \frac{1}{2} \right ) & = \sin \frac{5 \pi}{3}\end{align*} This is true because \begin{align*}\sin 300^\circ\end{align*} is \begin{align*}- \frac{\sqrt{3}}{2}\end{align*}
10. Change everything into terms of sine and cosine and simplify. \begin{align*}\sec x \cot x & = \csc x \\ \frac{1}{\cos x} \cdot \frac{\cos x}{\sin x} & = \frac{1}{\sin x} \\ \frac{1}{\sin x} & = \frac{1}{\sin x}\end{align*}

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