Verifying Trigonometric Identities Lesson

In this lesson, we will learn how to verify trigonometric identities, a skill that is essential for success in Calculus. In our last lesson, we reviewed the fundamental identities. Here, we will use those identities to verify that a trigonometric equation is an identity for those values of the variable for which it is defined.

Tips for Verifying Trigonometric Identities

1. Memorize the fundamental identities
2. Always try to work on the more complicated side
3. Express everything in terms of sine and cosine
4. Perform any indicated algebraic operations such as factoring or expanding
5. Use the formula for multiplying conjugates:
   • (a + b)(a - b)
   • (1 + sin x)(1 - sin x)
   • (1 + cos x)(1 - cos x)

Note: We are not solving an equation. Don't try things such as adding the same term to both sides or multiplying both sides by the same number. When working with identities, we work on one side at a time and try to rewrite such that it will match the other side. Let's look at some examples.
Example #1: Verify each identity. $$\frac{\text{cos}\hspace{.1em}x + \text{cot}\hspace{.1em}x}{\text{cot}\hspace{.1em}x}=1 + \text{sin}\hspace{.1em}x$$ In this case, the right side is less complicated, we will flip this to the left side for formatting: $$1 + \text{sin}\hspace{.1em}x=\frac{\text{cos}\hspace{.1em}x + \text{cot}\hspace{.1em}x}{\text{cot}\hspace{.1em}x}$$ Let's express the right side in terms of sine and cosine. $$1 + \text{sin}\hspace{.1em}x=\frac{\text{cos}\hspace{.1em}x + \text{cot}\hspace{.1em}x}{\text{cot}\hspace{.1em}x}$$ $$\hspace{.2em}=\frac{\text{cos}\hspace{.1em}x + \large{\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}}}{\large{\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}}}$$ Simplify: $$\hspace{.2em}=\frac{\text{cos}\hspace{.1em}x + \large{\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}}}{\large{\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}}}\cdot \frac{\text{sin}x}{\text{sin}x}$$ $$\hspace{.2em}=\frac{\text{sin}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x + \text{cos}\hspace{.1em}x}{\text{cos}\hspace{.1em}x}$$ Factor out the cos x and cancel: $$\require{cancel}\hspace{.2em}=\frac{\cancel{\text{cos}\hspace{.1em}x}(\text{sin}\hspace{.1em}x + 1)}{\cancel{\text{cos}\hspace{.1em}x}}$$ $$=1 + \text{sin}\hspace{.1em}x ✓$$ Example #2: Verify each identity. $$\frac{\text{cot}\hspace{.1em}x}{\text{sin}^2 \hspace{.1em}x}=\frac{\text{csc}^2 x}{\text{tan}\hspace{.1em}x}$$ In this case, each side is very similar. We notice that the left side contains sine, so we will work on the right side: $$\frac{\text{cot}\hspace{.1em}x}{\text{sin}^2 \hspace{.1em}x}=\frac{\text{csc}^2 x}{\text{tan}\hspace{.1em}x}$$ Let's write the right side in terms of sine and cosine: $$\hspace{.1em}=\Large{\frac{\frac{1}{\text{sin}^2 x}}{\frac{\text{sin}\hspace{.1em}x}{\text{cos}\hspace{.1em}x}}}$$ Simplify: $$\hspace{.1em}=\frac{1}{\text{sin}^2 x}\cdot \frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}$$ Replace with cot x: $$\hspace{.1em}=\frac{\text{cot}\hspace{.1em}x}{\text{sin}^2 x}✓$$ Example #3: Verify each identity. $$-\text{csc}^2 θ \hspace{.1em}\text{cos}^2 θ=1 - \text{csc}^2 θ$$ In this case, the right side is less complicated, we will flip this to the left side for formatting: $$1 - \text{csc}^2 θ=-\text{csc}^2 θ \hspace{.1em}\text{cos}^2 θ$$ Let's write the right side in terms of sine and cosine: $$\hspace{.1em}=-\frac{\text{cos}^2 θ}{\text{sin}^2 θ}$$ Replace with cot θ: $$\hspace{.1em}=-\text{cot}^2 θ$$ Use the Pythagorean Identity: $$\hspace{.1em}=1 - \text{csc}^2 θ✓$$ Example #4: Verify each identity. $$2 \text{tan}\hspace{.1em}θ \hspace{.1em}\text{sec}\hspace{.1em}θ=\frac{1}{1 - \text{sin}\hspace{.1em}θ}- \frac{1}{1 + \text{sin}\hspace{.1em}θ}$$ In this case, the right side is more complicated, so we will work with this side: $$2 \text{tan}\hspace{.1em}θ \hspace{.1em}\text{sec}\hspace{.1em}θ=\frac{1}{1 - \text{sin}\hspace{.1em}θ}- \frac{1}{1 + \text{sin}\hspace{.1em}θ}$$ Let's use the formula for conjugates: $$(a + b)(a - b)=a^2 - b^2$$ $$\hspace{.1em}=\frac{1}{1 - \text{sin}\hspace{.1em}θ}\cdot \frac{1 + \text{sin}\hspace{.1em}θ}{1 + \text{sin}\hspace{.1em}θ}- \frac{1}{1 + \text{sin}\hspace{.1em}θ}\cdot \frac{1 - \text{sin}\hspace{.1em}θ}{1 - \text{sin}\hspace{.1em}θ}$$ Simplify: $$\hspace{.1em}=\frac{1 + \text{sin}\hspace{.1em}θ}{1 - \text{sin}^2 θ}- \frac{1 - \text{sin}\hspace{.1em}θ}{1 - \text{sin}^2 θ}$$ Write with a common denominator: $$\hspace{.1em}=\frac{1 + \text{sin}\hspace{.1em}θ - (1 - \text{sin}\hspace{.1em}θ)}{1 - \text{sin}^2 θ}$$ Simplify: $$\hspace{.1em}=\frac{1 + \text{sin}\hspace{.1em}θ - 1 + \text{sin}\hspace{.1em}θ}{1 - \text{sin}^2 θ}$$ $$\hspace{.1em}=\frac{2\text{sin}\hspace{.1em}θ}{1 - \text{sin}^2 θ}$$ Use the Pythagorean Identity: $$\hspace{.1em}=\frac{2\text{sin}\hspace{.1em}θ}{\text{cos}^2 θ}$$ Let's break things up using fractions: $$\hspace{.1em}=2 \cdot \frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}\cdot \frac{1}{\text{cos}\hspace{.1em}θ}$$ Replace using tan θ and sec θ: $$\hspace{.1em}=2\text{tan}\hspace{.1em}θ \hspace{.1em}\text{sec}\hspace{.1em}θ✓$$

Verifying an Identity by Working Both Sides

In some cases, we need to find a common trigonometric expression. We will simplify one side, then flip and simplify the other. Let's look at an example.
Example #5: Verify each identity. $$\frac{\text{sec}\hspace{.1em}x + \text{tan}\hspace{.1em}x}{\text{sec}\hspace{.1em}x - \text{tan}\hspace{.1em}x}=\frac{1 + 2 \hspace{.1em}\text{sin}\hspace{.1em}x + \text{sin}^2 x}{\text{cos}^2 x}$$ Let's begin by working on just the left hand side. $$\frac{\text{sec}\hspace{.1em}x + \text{tan}\hspace{.1em}x}{\text{sec}\hspace{.1em}x - \text{tan}\hspace{.1em}x}$$ Since cos x multiplied by sec x is 1, we will multiply by cos x over cos x: $$\hspace{.1em}=\frac{\text{sec}\hspace{.1em}x + \text{tan}\hspace{.1em}x}{\text{sec}\hspace{.1em}x - \text{tan}\hspace{.1em}x}\cdot \frac{\text{cos}\hspace{.1em}x}{\text{cos}\hspace{.1em}x}$$ $$\hspace{.1em}=\frac{\text{sec}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x + \text{tan}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x}{\text{sec}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x - \text{tan}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x}$$ Replace sec x cos x with 1: $$\hspace{.1em}=\frac{1 + \text{tan}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x}{1 - \text{tan}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x}$$ Let's convert tangent into sine and cosine: $$\hspace{.1em}=\frac{1 + \large{\frac{\text{sin}\hspace{.1em}x}{\text{cos}\hspace{.1em}x}}\hspace{.1em}\text{cos}\hspace{.1em}x}{1 - \large{\frac{\text{sin}\hspace{.1em}x}{\text{cos}\hspace{.1em}x}}\hspace{.1em}\text{cos}\hspace{.1em}x}$$ Simplify: $$\hspace{.1em}=\frac{1 + \text{sin}\hspace{.1em}x}{1 - \text{sin}\hspace{.1em}x}$$ Now, let's switch over and work on the right hand side. $$\frac{1 + 2 \hspace{.1em}\text{sin}\hspace{.1em}x + \text{sin}^2 x}{\text{cos}^2 x}$$ Here, we can factor the numerator: $$x^2 + 2xy + y^2=(x + y)^2$$ $$\hspace{.1em}=\frac{(1 + \text{sin}\hspace{.1em}x)^2}{\text{cos}^2 x}$$ Use the Pythagorean Identity: $$\hspace{.1em}=\frac{(1 + \text{sin}\hspace{.1em}x)^2}{1 - \text{sin}^2 x}$$ Factor the denominator: $$a^2 - b^2=(a + b)(a - b)$$ $$=\frac{(1 + \text{sin}\hspace{.1em}x)^2}{(1 - \text{sin}\hspace{.1em}x)(1 + \text{sin}\hspace{.1em}x)}$$ Cancel Common Factors: $$\hspace{.1em}=\frac{1 + \text{sin}\hspace{.1em}x}{1 - \text{sin}\hspace{.1em}x}$$ We have shown that the left and right sides are equal to a common trigonometric expression: $$\frac{\text{sec}\hspace{.1em}x + \text{tan}\hspace{.1em}x}{\text{sec}\hspace{.1em}x - \text{tan}\hspace{.1em}x}=\frac{1 + \text{sin}\hspace{.1em}x}{1 - \text{sin}\hspace{.1em}x}=\frac{1 + 2 \hspace{.1em}\text{sin}\hspace{.1em}x + \text{sin}^2 x}{\text{cos}^2 x}✓$$

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