Fundamental Identities Lesson

Negative-Angle Identities

Let's begin our lesson by thinking about Negative-Angle Identities. If we look at our image above, we can see that an angle θ that has the point (x,y) on its terminal side will have a corresponding angle -θ with the point (x, -y) on its terminal side. If we think about this using the definition of sine: $$\text{sin}\hspace{.15em}θ=\frac{y}{r}$$ $$\text{sin}\hspace{.15em}(-θ)=\frac{-y}{r}=-\frac{y}{r}$$ This tells us that sin(-θ) and sin θ are negatives of each other. This gives us our first Negative-Angle Identity: $$\text{sin}(-θ)=-\text{sin}\hspace{.15em}θ$$ Additionally, we can use the image to show: $$\text{cos}\hspace{.15em}θ=\frac{x}{r}$$ $$\text{cos}(-θ)=\frac{x}{r}$$ This gives us our second Negative-Angle Identity: $$\text{cos}(-θ)=\text{cos}\hspace{.15em}θ$$ Lastly, we can use the identities for sin(-θ) and cos(-θ) to find tan(-θ) in terms of tan θ: Scroll Right for More: $$\text{tan}(-θ)=\frac{\text{sin}(-θ)}{\text{cos}(-θ)}=\frac{-\text{sin}\hspace{.15em}θ}{\text{cos}\hspace{.15em}θ}=-\frac{\text{sin}\hspace{.15em}θ}{\text{cos}\hspace{.15em}θ}$$ This gives us our third Negative-Angle Identity: $$\text{tan}(-θ)=-\text{tan}\hspace{.15em}θ$$ Using a similar thought process, we can obtain our other three Negative-Angle Identities: $$\text{csc}(-θ)=-\text{csc}\hspace{.15em}θ$$ $$\text{sec}(-θ)=\text{sec}\hspace{.15em}θ$$ $$\text{cot}(-θ)=-\text{cot}\hspace{.15em}θ$$

Fundamental Identities

We previously learned about the Reciprocal Identities, Pythagorean Identities, and Quotient Identities. When added to the Negative-Angle Identities, we obtain the Fundamental Identities.

Reciprocal Identities
sin θ	=	$\frac{1}{\text{csc}\hspace{.15em}θ}$
csc θ	=	$\frac{1}{\text{sin}\hspace{.15em}θ}$
cos θ	=	$\frac{1}{\text{sec}\hspace{.15em}θ}$
sec θ	=	$\frac{1}{\text{cos}\hspace{.15em}θ}$
tan θ	=	$\frac{1}{\text{cot}\hspace{.15em}θ}$
cot θ	=	$\frac{1}{\text{tan}\hspace{.15em}θ}$
Pythagorean Identities
sin2 θ + cos2θ	=	1
tan2 θ + 1	=	sec2θ
1 + cot2 θ	=	csc2θ
Quotient Identities
tan θ	=	$\frac{\text{sin}\hspace{.15em}θ}{\text{cos}\hspace{.15em}θ}$
cot θ	=	$\frac{\text{cos}\hspace{.15em}θ}{\text{sin}\hspace{.15em}θ}$
Negative-Angle Identities
sin(-θ)	=	-sin θ
cos(-θ)	=	cos θ
tan(-θ)	=	-tan θ
csc(-θ)	=	-csc θ
sec(-θ)	=	sec θ
cot(-θ)	=	-cot θ

Using the Fundamental Identities

We will sometimes be asked to find the values of other trigonometric functions from the value of a given trigonometric function. When using the fundamental identities to solve these problems, we want to use an identity that relates the function we have to the function we want. Let's look at an example.
Example #1: Find each function value. $$\text{tan}\hspace{.15em}θ=\frac{3}{4}$$ $$\text{sin}\hspace{.15em}θ > 0$$ Find sec θ, sin θ, and cot(-θ)
In this case, we have tangent and need secant. For this situation, we can use the Pythagorean Identity: $$\text{tan}^2 \hspace{.15em}θ + 1=\text{sec}^2 θ$$ Let's plug in for tan θ: $$\left(\frac{3}{4}\right)^2 + 1=\text{sec}^2 θ$$ $$\frac{9}{16}+ 1=\text{sec}^2 θ$$ $$\frac{9}{16}+ \frac{16}{16}=\text{sec}^2 θ$$ $$\frac{25}{16}=\text{sec}^2 θ$$ $$\text{sec}^2 θ=\frac{25}{16}$$ $$\text{sec}\hspace{.15em}θ=\pm\sqrt{\frac{25}{16}}$$ Since sine is positive and tangent is positive, we know we are in quadrant I, where everything is positive. This means we can throw out the negative and keep the principal square root. $$\text{sec}\hspace{.15em}θ=\frac{5}{4}$$ For the second part, we are asked to find sin θ. Since we already know the sec θ, we can use our Quotient Identity: $$\text{tan}\hspace{.15em}θ=\frac{\text{sin}\hspace{.15em}θ}{\text{cos}\hspace{.15em}θ}$$ We can replace cos θ with $\frac{1}{\text{sec}\hspace{.15em}θ}$: $$\text{cos}\hspace{.15em}θ=\frac{1}{\text{sec}\hspace{.15em}θ}$$ $$\text{cos}\hspace{.15em}θ=\frac{1}{\large\frac{5}{4}}=\frac{4}{5}$$ We can rewrite our Quotient Identity: $$\text{sin}\hspace{.15em}θ=\text{tan}\hspace{.15em}θ \cdot \text{cos}\hspace{.15em}θ$$ Now, we can plug in for tan θ and cos θ: $$\require{cancel}\text{sin}\hspace{.15em}θ=\frac{3}{\cancel{4}}\cdot \frac{\cancel{4}}{5}=\frac{3}{5}$$ For the final part, we are asked to find cot(-θ). First, let's start with our Reciprocal Identity: $$\text{cot}\hspace{.15em}(-θ)=\frac{1}{\text{tan}(-θ)}$$ Now, let's apply our Negative-Angle Identity: $$\text{cot}\hspace{.15em}(-θ)=\frac{1}{-\text{tan}\hspace{.15em}θ}$$ Now, we can plug in for tan θ: $$\text{cot}\hspace{.15em}(-θ)=\frac{1}{\large{-\frac{3}{4}}}=-\frac{4}{3}$$

Writing One Trigonometric Function in Terms of Another

In some cases, we will be asked to use the fundamental identities to write one trigonometric function in terms of another. Let's look at an example.
Example #2: Write cos θ in terms of tan θ.
First, let's think about the Pythagorean Identity that includes tan θ and sec θ: $$1 + \text{tan}^2 θ=\text{sec}^2 θ$$ Let's now think about our Reciprocal Identity that includes cos θ and sec θ: $$\text{cos}\hspace{.15em}θ=\frac{1}{\text{sec}\hspace{.15em}θ}$$ Let's take the reciprocal of each side: $$\frac{1}{1 + \text{tan}^2 θ}=\frac{1}{\text{sec}^2 θ}$$ Using our Reciprocal Identity above, we can now rewrite the right side: $$\frac{1}{1 + \text{tan}^2 θ}=\text{cos}^2 θ$$ Now, we want to take the square root of each side: $$\text{cos}\hspace{.15em}θ=\pm \sqrt{\frac{1}{1 + \text{tan}^2 θ}}$$ Let's simplify and rationalize the denominator: $$\text{cos}\hspace{.15em}θ=\frac{\pm 1}{\sqrt{1 + \text{tan}^2 θ}}$$ $$\text{cos}\hspace{.15em}θ=\frac{\pm \sqrt{1 + \text{tan}^2 θ}}{1 + \text{tan}^2 θ}$$

Rewriting an Expression in Terms of Sine and Cosine

We are often asked to simplify trigonometric expressions by writing our expression in terms of sine and cosine. Let's look at an example.
Example #3: Simplify each. $$\text{sec}\hspace{.15em}θ\cdot \text{cot}\hspace{.15em}θ \cdot \text{sin}\hspace{.15em}θ$$ Let's replace our sec θ and cot θ using the reciprocal identities. This will put our expression in terms of sine and cosine. $$\text{sec}\hspace{.15em}θ=\frac{1}{\text{cos}\hspace{.15em}θ}$$ $$\text{cot}\hspace{.15em}θ=\frac{\text{cos}\hspace{.15em}θ}{\text{sin}\hspace{.15em}θ}$$ $$\frac{1}{\text{cos}\hspace{.15em}θ}\cdot \frac{\text{cos}\hspace{.15em}θ}{\text{sin}\hspace{.15em}θ}\cdot \text{sin}\hspace{.15em}θ$$ Let's cancel common factors: $$\frac{1}{\cancel{\text{cos}\hspace{.15em}θ}}\cdot \frac{\cancel{\text{cos}\hspace{.15em}θ}}{\cancel{\text{sin}\hspace{.15em}θ}}\cdot \cancel{\text{sin}\hspace{.15em}θ}=1$$ Example #4: Simplify each. $$\text{sin}^2(-θ) + \text{tan}^2(-θ) + \text{cos}^2(-θ)$$ For this problem, we need to be really careful with our signs. Recall that: $$(\text{sin}\hspace{.15em}θ)^2=\text{sin}^2 θ$$ Let's rewrite our problem as such: Scroll Right for More: $$(\text{sin}\hspace{.1em}({-}θ))^2 + (\text{tan}({-}θ))^2 + (\text{cos}({-}θ))^2$$ Now, let's use our Negative-Angle identities: $$(-\text{sin}\hspace{.1em}θ)^2 + (-\text{tan}\hspace{.1em}θ)^2 + (\text{cos}\hspace{.1em}θ)^2$$ When we square a number, the result is non-negative. This allows us to drop the negative signs in front of sin θ and tan θ: $$(\text{sin}\hspace{.1em}θ)^2 + (\text{tan}\hspace{.1em}θ)^2 + (\text{cos}\hspace{.1em}θ)^2$$ Now, let's revert back to our original form: $$\text{sin}^2 \hspace{.1em}θ + \text{tan}^2 \hspace{.1em}θ + \text{cos}^2 \hspace{.1em}θ$$ Let's rearrange our expression: $$\text{sin}^2 \hspace{.1em}θ + \text{cos}^2 \hspace{.1em}θ + \text{tan}^2 \hspace{.1em}θ$$ From our Pythagorean Identities: $$\text{sin}^2 \hspace{.1em}θ + \text{cos}^2 \hspace{.1em}θ=1$$ Let's replace this in our expression: $$1 + \text{tan}^2 \hspace{.1em}θ$$ From our Pythagorean Identities: $$\text{tan}^2 \hspace{.1em}θ + 1=\text{sec}^2 \hspace{.1em}θ$$ Let's replace this in our expression: $$\text{sec}^2 \hspace{.1em}θ$$

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