Lesson Objectives
- Learn how to solve problems using the Pythagorean Identities
- Learn how to solve problems using the Quotient Identities
Pythagorean & Quotient Identities, Finding Missing Function Values
In our last lesson, we learned about the Reciprocal Identities. Now, we will move on and discuss the Pythagorean and Quotient Identities. For reference, let's list the six trigonometric functions: $$\text{sin}\hspace{.25em}θ=\frac{y}{r}$$ $$\text{cos}\hspace{.25em}θ=\frac{x}{r}$$ $$\text{tan}\hspace{.25em}θ=\frac{y}{x}, x ≠ 0$$ $$\text{csc}\hspace{.25em}θ=\frac{r}{y}, y ≠ 0$$ $$\text{sec}\hspace{.25em}θ=\frac{r}{x}, x ≠ 0$$ $$\text{cot}\hspace{.25em}θ=\frac{x}{y}, y ≠ 0$$
Let's look at a few examples.
Example #1: Find cos θ and sin θ.
$$\text{tan}\hspace{.2em}θ=\frac{7}{3}$$ $$\text{sin}\hspace{.2em}θ < 0$$ A common mistake is to deduce that since tan θ is sin θ/cos θ, that sin θ is -7 and cos θ is -3. This is not correct since sin θ and cos θ must be in the interval [-1, 1].
Since we are given the value for tan θ, let's plug into the appropriate Pythagorean Identity: $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ Let's plug in for tan θ: $$\left(\frac{7}{3}\right)^2 + 1=\text{sec}^2 \hspace{.2em}θ$$ $$\frac{49}{9}+ 1=\text{sec}^2 \hspace{.2em}θ$$ On the right side, let's get a common denominator and find our sum: $$\frac{49}{9}+ \frac{9}{9}=\text{sec}^2 \hspace{.2em}θ$$ $$\frac{58}{9}=\text{sec}^2 \hspace{.2em}θ$$ To solve for sec θ, let's take the square root of each side: $$\text{sec}\hspace{.2em}θ=\pm \sqrt{\frac{58}{9}}=\pm \frac{\sqrt{58}}{3}$$ Since we know that tan θ is positive and sin θ is negative, we know that our angle is in quadrant III. This means that sec θ is negative. We will only use the negative square root as our answer: $$\text{sec}\hspace{.2em}θ=- \frac{\sqrt{58}}{3}$$ Since sec θ is the reciprocal of cos θ, we can flip our answer and we have cos θ: $$\text{cos}\hspace{.2em}θ=-\frac{3}{\sqrt{58}}$$ Let's rationalize our denominator: $$\text{cos}\hspace{.2em}θ=-\frac{3 \sqrt{58}}{58}$$ To find sin θ, we can plug into our other Pythagorean Identity: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ Let's plug in for cos θ: $$\text{sin}^2 \hspace{.2em}θ + \left(-\frac{3}{\sqrt{58}}\right)^2=1$$ $$\text{sin}^2 \hspace{.2em}θ + \frac{9}{58}=1$$ Let's subtract 9/58 away from each side: $$\text{sin}^2 \hspace{.2em}θ=1 - \frac{9}{58}$$ Let's get a common denominator and subtract: $$\text{sin}^2 \hspace{.2em}θ=\frac{58}{58}- \frac{9}{58}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{58 - 9}{58}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{49}{58}$$ To get this in terms of sin θ, we will take the square root of each side. We will take the negative square root of the right side since we know that sin θ is negative in quadrant III. $$\text{sin}\hspace{.2em}θ=-\sqrt{\frac{49}{58}}$$ $$\text{sin}\hspace{.2em}θ=-\frac{7}{\sqrt{58}}$$ Let's rationalize our denominator: $$\text{sin}\hspace{.2em}θ=-\frac{7 \sqrt{58}}{58}$$ Example #2: Find sin θ and tan θ.
$$\text{sec}\hspace{.2em}θ=-2$$ $$\text{cot}\hspace{.2em}θ > 0$$ Since we are given the value for sec θ, let's plug into the appropriate Pythagorean Identity: $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ Let's plug in for sec θ: $$\text{tan}^2 \hspace{.2em}θ + 1=(-2)^2$$ $$\text{tan}^2 \hspace{.2em}θ + 1=4$$ Subtract 1 away from each side: $$\text{tan}^2 \hspace{.2em}θ=3$$ Let's get tan θ by taking the square root of each side. Since sec θ is negative and cot θ is positive, this means we are in quadrant III. We will take the principal square root of the right side since tan θ is positive in quadrant III. $$\text{tan}^2 \hspace{.2em}θ=3$$ $$\text{tan}\hspace{.2em}θ=\sqrt{3}$$ Now that we know tan θ, how can we find sin θ? We can use our other Pythagorean Identity: $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$ We know that tan θ and cot θ are reciprocals: $$\text{cot}\hspace{.2em}θ=\frac{1}{\text{tan}\hspace{.2em}θ}$$ $$\text{cot}\hspace{.2em}θ=\frac{1}{\sqrt{3}}$$ Now, let's plug in for cot θ: $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$ $$1 + \left(\frac{1}{\sqrt{3}}\right)^2=\text{csc}^2 \hspace{.2em}θ$$ $$1 + \frac{1}{3}=\text{csc}^2 \hspace{.2em}θ$$ Let's get a common denominator and simplify the right side: $$\frac{3}{3}+ \frac{1}{3}=\text{csc}^2 \hspace{.2em}θ$$ $$\frac{4}{3}=\text{csc}^2 \hspace{.2em}θ$$ $$\text{csc}^2 \hspace{.2em}θ=\frac{4}{3}$$ Let's take the square root of each side. Since csc θ is negative in quadrant III, let's use the negative square root on the right side: $$\text{csc}\hspace{.2em}θ=-\sqrt{\frac{4}{3}}$$ $$\text{csc}\hspace{.2em}θ=-\frac{2}{\sqrt{3}}$$ Here, we don't need to rationalize the denominator since we want sin θ, which is the reciprocal of csc θ: $$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{3}}{2}$$ Example #3: Find sin θ and tan θ.
$$\text{cos}\hspace{.2em}θ=-\frac{2\sqrt{2}}{3}$$ $$\text{sin}\hspace{.2em}θ < 0$$ Since we are given the value for cos θ, let's plug into the appropriate Pythagorean Identity: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ Let's plug in for cos θ: $$\text{sin}^2 \hspace{.2em}θ + \left(-\frac{2\sqrt{2}}{3}\right)^2=1$$ $$\text{sin}^2 \hspace{.2em}θ + \frac{8}{9}=1$$ Subtract 8/9 away from each side: $$\text{sin}^2 \hspace{.2em}θ=1 - \frac{8}{9}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{9}{9}- \frac{8}{9}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{1}{9}$$ To get sin θ by itself, let's take the square root of each side. Since sin θ is negative, we will use the negative square root on the right side: $$\text{sin}\hspace{.2em}θ=-\sqrt{\frac{1}{9}}$$ $$\text{sin}\hspace{.2em}θ=-\frac{1}{3}$$ Now to find tan θ we can use one of our Quotient Identities: $$\frac{\text{sin}\hspace{.2em}θ}{\text{cos}\hspace{.2em}θ}=\text{tan}\hspace{.2em}θ$$ Plug in for sin θ and cos θ: $$\large{\frac{-\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}}=\text{tan}\hspace{.2em}θ$$ $$\text{tan}\hspace{.2em}θ=-\frac{1}{3}\cdot -\frac{3}{2 \sqrt{2}}$$ $$\text{tan}\hspace{.2em}θ=-\frac{1}{\cancel{3}}\cdot -\frac{\cancel{3}}{2 \sqrt{2}}$$ $$\text{tan}\hspace{.2em}θ=\frac{1}{2\sqrt{2}}$$ Let's rationalize our denominator: $$\text{tan}\hspace{.2em}θ=\frac{\sqrt{2}}{4}$$
Pythagorean Identities
Let's begin by writing our Pythagorean formula: $$a^2 + b^2=c^2$$ In our case, x and y are the legs of our right triangle and represent a and b. Let's replace this in our formula: $$x^2 + y^2=c^2$$ Since r is our hypotenuse, it represents c. Let's replace this in our formula: $$x^2 + y^2=r^2$$ First, let's divide everything by r2: $$\frac{x^2}{r^2}+ \frac{y^2}{r^2}=\frac{r^2}{r^2}$$ Since r2/r2 is 1, we can replace this: $$\frac{x^2}{r^2}+ \frac{y^2}{r^2}=1$$ Additionally, we can use our rules of exponents to rewrite our equation: $$\left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2=1$$ At this point, we can replace x/r with cos θ and y/r with sin θ: $$(\text{cos}\hspace{.2em}θ)^2 + (\text{sin}\hspace{.2em}θ)^2=1$$ Now, we can apply our exponents and use the commutative property of addition to switch the order: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ We can use a similar process to find the other two Pythagorean Identities. If we divide everything by x2, we will obtain: $$x^2 + y^2=r^2$$ $$\frac{x^2}{x^2}+ \frac{y^2}{x^2}=\frac{r^2}{x^2}$$ $$1 + \left(\frac{y}{x}\right)^2=\left(\frac{r}{x}\right)^2$$ $$1 + \text{tan}^2 \hspace{.2em}θ=\text{sec}^2 \hspace{.2em}θ$$ Generally, this is rearranged using the commutative property of addition: $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ Lastly, if we divide everything by y2, we will obtain: $$x^2 + y^2=r^2$$ $$\frac{x^2}{y^2}+ \frac{y^2}{y^2}=\frac{r^2}{y^2}$$ $$\left(\frac{x}{y}\right)^2 + 1=\left(\frac{r}{y}\right)^2$$ $$\text{cot}^2 \hspace{.2em}θ + 1=\text{csc}^2 \hspace{.2em}θ$$ Generally, this is rearranged using the commutative property of addition: $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$ Pythagorean Identities: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$Quotient Identities
Additionally, we have the following Quotient Identities: $$\frac{\text{sin}\hspace{.2em}θ}{\text{cos}\hspace{.2em}θ}=\text{tan}\hspace{.2em}θ$$ $$\require{cancel}\frac{y}{r}\div \frac{x}{r}=\frac{y}{\cancel{r}}\cdot \frac{\cancel{r}}{x}=\frac{y}{x}$$ $$\frac{\text{cos}\hspace{.2em}θ}{\text{sin}\hspace{.2em}θ}=\text{cot}\hspace{.2em}θ$$ $$\frac{x}{r}\div \frac{y}{r}=\frac{x}{\cancel{r}}\cdot \frac{\cancel{r}}{y}=\frac{x}{y}$$Ranges of Trigonometric Functions
In many cases, knowing the ranges of the six trigonometric functions will be extremely useful. Keep these ranges in mind when solving problems:| Trig Function | Range |
|---|---|
| sin θ, cos θ | [-1, 1] |
| tan θ, cot θ | (-∞, ∞) |
| sec θ, csc θ | (-∞, -1] ∪ [1, ∞) |
Example #1: Find cos θ and sin θ.
$$\text{tan}\hspace{.2em}θ=\frac{7}{3}$$ $$\text{sin}\hspace{.2em}θ < 0$$ A common mistake is to deduce that since tan θ is sin θ/cos θ, that sin θ is -7 and cos θ is -3. This is not correct since sin θ and cos θ must be in the interval [-1, 1].
Since we are given the value for tan θ, let's plug into the appropriate Pythagorean Identity: $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ Let's plug in for tan θ: $$\left(\frac{7}{3}\right)^2 + 1=\text{sec}^2 \hspace{.2em}θ$$ $$\frac{49}{9}+ 1=\text{sec}^2 \hspace{.2em}θ$$ On the right side, let's get a common denominator and find our sum: $$\frac{49}{9}+ \frac{9}{9}=\text{sec}^2 \hspace{.2em}θ$$ $$\frac{58}{9}=\text{sec}^2 \hspace{.2em}θ$$ To solve for sec θ, let's take the square root of each side: $$\text{sec}\hspace{.2em}θ=\pm \sqrt{\frac{58}{9}}=\pm \frac{\sqrt{58}}{3}$$ Since we know that tan θ is positive and sin θ is negative, we know that our angle is in quadrant III. This means that sec θ is negative. We will only use the negative square root as our answer: $$\text{sec}\hspace{.2em}θ=- \frac{\sqrt{58}}{3}$$ Since sec θ is the reciprocal of cos θ, we can flip our answer and we have cos θ: $$\text{cos}\hspace{.2em}θ=-\frac{3}{\sqrt{58}}$$ Let's rationalize our denominator: $$\text{cos}\hspace{.2em}θ=-\frac{3 \sqrt{58}}{58}$$ To find sin θ, we can plug into our other Pythagorean Identity: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ Let's plug in for cos θ: $$\text{sin}^2 \hspace{.2em}θ + \left(-\frac{3}{\sqrt{58}}\right)^2=1$$ $$\text{sin}^2 \hspace{.2em}θ + \frac{9}{58}=1$$ Let's subtract 9/58 away from each side: $$\text{sin}^2 \hspace{.2em}θ=1 - \frac{9}{58}$$ Let's get a common denominator and subtract: $$\text{sin}^2 \hspace{.2em}θ=\frac{58}{58}- \frac{9}{58}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{58 - 9}{58}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{49}{58}$$ To get this in terms of sin θ, we will take the square root of each side. We will take the negative square root of the right side since we know that sin θ is negative in quadrant III. $$\text{sin}\hspace{.2em}θ=-\sqrt{\frac{49}{58}}$$ $$\text{sin}\hspace{.2em}θ=-\frac{7}{\sqrt{58}}$$ Let's rationalize our denominator: $$\text{sin}\hspace{.2em}θ=-\frac{7 \sqrt{58}}{58}$$ Example #2: Find sin θ and tan θ.
$$\text{sec}\hspace{.2em}θ=-2$$ $$\text{cot}\hspace{.2em}θ > 0$$ Since we are given the value for sec θ, let's plug into the appropriate Pythagorean Identity: $$\text{tan}^2 \hspace{.2em}θ + 1=\text{sec}^2 \hspace{.2em}θ$$ Let's plug in for sec θ: $$\text{tan}^2 \hspace{.2em}θ + 1=(-2)^2$$ $$\text{tan}^2 \hspace{.2em}θ + 1=4$$ Subtract 1 away from each side: $$\text{tan}^2 \hspace{.2em}θ=3$$ Let's get tan θ by taking the square root of each side. Since sec θ is negative and cot θ is positive, this means we are in quadrant III. We will take the principal square root of the right side since tan θ is positive in quadrant III. $$\text{tan}^2 \hspace{.2em}θ=3$$ $$\text{tan}\hspace{.2em}θ=\sqrt{3}$$ Now that we know tan θ, how can we find sin θ? We can use our other Pythagorean Identity: $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$ We know that tan θ and cot θ are reciprocals: $$\text{cot}\hspace{.2em}θ=\frac{1}{\text{tan}\hspace{.2em}θ}$$ $$\text{cot}\hspace{.2em}θ=\frac{1}{\sqrt{3}}$$ Now, let's plug in for cot θ: $$1 + \text{cot}^2 \hspace{.2em}θ=\text{csc}^2 \hspace{.2em}θ$$ $$1 + \left(\frac{1}{\sqrt{3}}\right)^2=\text{csc}^2 \hspace{.2em}θ$$ $$1 + \frac{1}{3}=\text{csc}^2 \hspace{.2em}θ$$ Let's get a common denominator and simplify the right side: $$\frac{3}{3}+ \frac{1}{3}=\text{csc}^2 \hspace{.2em}θ$$ $$\frac{4}{3}=\text{csc}^2 \hspace{.2em}θ$$ $$\text{csc}^2 \hspace{.2em}θ=\frac{4}{3}$$ Let's take the square root of each side. Since csc θ is negative in quadrant III, let's use the negative square root on the right side: $$\text{csc}\hspace{.2em}θ=-\sqrt{\frac{4}{3}}$$ $$\text{csc}\hspace{.2em}θ=-\frac{2}{\sqrt{3}}$$ Here, we don't need to rationalize the denominator since we want sin θ, which is the reciprocal of csc θ: $$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{3}}{2}$$ Example #3: Find sin θ and tan θ.
$$\text{cos}\hspace{.2em}θ=-\frac{2\sqrt{2}}{3}$$ $$\text{sin}\hspace{.2em}θ < 0$$ Since we are given the value for cos θ, let's plug into the appropriate Pythagorean Identity: $$\text{sin}^2 \hspace{.2em}θ + \text{cos}^2 \hspace{.2em}θ=1$$ Let's plug in for cos θ: $$\text{sin}^2 \hspace{.2em}θ + \left(-\frac{2\sqrt{2}}{3}\right)^2=1$$ $$\text{sin}^2 \hspace{.2em}θ + \frac{8}{9}=1$$ Subtract 8/9 away from each side: $$\text{sin}^2 \hspace{.2em}θ=1 - \frac{8}{9}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{9}{9}- \frac{8}{9}$$ $$\text{sin}^2 \hspace{.2em}θ=\frac{1}{9}$$ To get sin θ by itself, let's take the square root of each side. Since sin θ is negative, we will use the negative square root on the right side: $$\text{sin}\hspace{.2em}θ=-\sqrt{\frac{1}{9}}$$ $$\text{sin}\hspace{.2em}θ=-\frac{1}{3}$$ Now to find tan θ we can use one of our Quotient Identities: $$\frac{\text{sin}\hspace{.2em}θ}{\text{cos}\hspace{.2em}θ}=\text{tan}\hspace{.2em}θ$$ Plug in for sin θ and cos θ: $$\large{\frac{-\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}}=\text{tan}\hspace{.2em}θ$$ $$\text{tan}\hspace{.2em}θ=-\frac{1}{3}\cdot -\frac{3}{2 \sqrt{2}}$$ $$\text{tan}\hspace{.2em}θ=-\frac{1}{\cancel{3}}\cdot -\frac{\cancel{3}}{2 \sqrt{2}}$$ $$\text{tan}\hspace{.2em}θ=\frac{1}{2\sqrt{2}}$$ Let's rationalize our denominator: $$\text{tan}\hspace{.2em}θ=\frac{\sqrt{2}}{4}$$
Skills Check:
Example #1
Find cos θ $$\text{sin}\hspace{.2em}θ=-\frac{1}{2}$$ $$\text{cot}\hspace{.2em}θ > 0$$
Please choose the best answer.
A
$$\text{cos}\hspace{.2em}θ=\frac{2\sqrt{3}}{3}$$
B
$$\text{cos}\hspace{.2em}θ=-2$$
C
$$\text{cos}\hspace{.2em}θ=-\frac{\sqrt{3}}{2}$$
D
$$\text{cos}\hspace{.2em}θ=-\sqrt{3}$$
E
$$\text{cos}\hspace{.2em}θ=-\frac{5}{2}$$
Example #2
Find cot θ $$\text{sec}\hspace{.2em}θ=-\frac{5}{2}$$ $$\text{cot}\hspace{.2em}θ < 0$$
Please choose the best answer.
A
$$\text{cot}\hspace{.2em}θ=\frac{\sqrt{7}}{21}$$
B
$$\text{cot}\hspace{.2em}θ=-\frac{5}{21}$$
C
$$\text{cot}\hspace{.2em}θ=-\frac{\sqrt{21}}{2}$$
D
$$\text{cot}\hspace{.2em}θ=\frac{5\sqrt{21}}{21}$$
E
$$\text{cot}\hspace{.2em}θ=-\frac{2\sqrt{21}}{21}$$
Example #3
Find sin θ $$\text{csc}\hspace{.2em}θ=\frac{4}{3}$$ $$\text{cos}\hspace{.2em}θ > 0$$
Please choose the best answer.
A
$$\text{sin}\hspace{.2em}θ=-\sqrt{2}$$
B
$$\text{sin}\hspace{.2em}θ=\frac{4}{3}$$
C
$$\text{sin}\hspace{.2em}θ=\frac{4\sqrt{7}}{7}$$
D
$$\text{sin}\hspace{.2em}θ=\frac{3}{4}$$
E
$$\text{sin}\hspace{.2em}θ=\frac{\sqrt{7}}{4}$$
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