Lesson Objectives
  • Learn how to solve problems using reciprocal identities
  • Learn how to find the signs of function values
  • Learn how to find the quadrant of an angle based on conditions
  • Learn how to find all function values given one value and the quadrant

Reciprocal Identities, Signs of Function Values, & Finding Missing Function Values


Identities

In our algebra course, we learned that identities are equations that are always true. For example: $$3(x + 5)=3x + 15$$ The equation above is an example of an identity, we can replace x with any real number, and we will always get a true statement. This type of equation has a solution that is said to be "all real numbers" or the interval: $$(-\infty, \infty)$$

Reciprocal Identities

The reciprocal of a nonzero number is found by flipping the number. In other words, the numerator becomes the denominator and the denominator becomes the numerator. $$\frac{3}{4}$$ The reciprocal of 3/4 is 4/3, we interchange the numerator and the denominator. $$\frac{4}{3}$$ In our last section, we introduced the trigonometric functions and found that some functions were reciprocals of each other. We know that we can't divide by zero, so keep in mind that anytime we have zero in the denominator, we write a result of undefined. $$r=\sqrt{x^2 + y^2}$$ $$\text{sin}\hspace{.25em}θ=\frac{y}{r}$$ $$\text{csc}\hspace{.25em}θ=\frac{r}{y}, y ≠ 0$$ $$\text{sin}\hspace{.25em}θ=\frac{1}{\text{csc}\hspace{.25em}θ}$$ $$\text{csc}\hspace{.25em}θ=\frac{1}{\text{sin}\hspace{.25em}θ}$$ We can see that sin θ and csc θ are reciprocals. $$\text{cos}\hspace{.25em}θ=\frac{x}{r}$$ $$\text{sec}\hspace{.25em}θ=\frac{r}{x}, x ≠ 0$$ $$\text{cos}\hspace{.25em}θ=\frac{1}{\text{sec}\hspace{.25em}θ}$$ $$\text{sec}\hspace{.25em}θ=\frac{1}{\text{cos}\hspace{.25em}θ}$$ Additionally, cos θ and sec θ are reciprocals. $$\text{tan}\hspace{.25em}θ=\frac{y}{x}, x ≠ 0$$ $$\text{cot}\hspace{.25em}θ=\frac{x}{y}, y ≠ 0$$ $$\text{tan}\hspace{.25em}θ=\frac{1}{\text{cot}\hspace{.25em}θ}$$ $$\text{cot}\hspace{.25em}θ=\frac{1}{\text{tan}\hspace{.25em}θ}$$ Lastly, tan θ and cot θ are reciprocals. Let's look at a few examples.
Example #1: Find each function value. $$\text{cos}\hspace{.2em}θ$$ $$\text{sec}\hspace{.2em}θ=\frac{9}{2}$$ Since cos θ and sec θ are reciprocals, we can just flip our fraction to obtain our answer. $$\text{cos}\hspace{.2em}θ=\frac{2}{9}$$ Example #2: Find each function value. $$\text{csc}\hspace{.2em}θ$$ $$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{6}}{3}$$ Since sin θ and csc θ are reciprocals, we can just flip our fraction to obtain our answer. Since this will lead to a radical in the denominator, we need to take the extra step and rationalize the denominator. $$\text{csc}\hspace{.2em}θ=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}$$

Signs of Function Values

When we defined the trigonometric functions, we gave r as the distance from the origin or the point (0,0) to the point (x,y). Since this distance is undirected, the value of r is always positive. When thinking about the values of our six trigonometric functions, we should remember that the value of x is positive in quadrants I and IV and negative in quadrants II and III. For y, the value is positive in quadrants I and II and negative in quadrants III and IV. Showing signs of (x,y) based on the quadrant Using these facts, we can produce a table for the signs of function values:
Quadrant sin θ cos θ tan θ cot θ sec θ csc θ
I++++++
II+----+
III--++--
IV-+--+-
Some students use the phrase "All Students Take Calculus" to remember the signs.
  1. All Functions Positive
  2. Students (Sine and Cosecant) Positive
  3. Take (Tangent and Cotangent) Positive
  4. Calculus (Cosine and Secant) Positive
Let's look at a few examples.
Example #3: Determine the signs of the trigonometric functions of an angle in standard position with the given measure. $$110°$$ A 110° angle that is in standard position will lie in quadrant II. The x-values will be negative and the y-values will be positive. Showing a 110 degree angle in standard position Using our chart above, we can see that sin θ, and csc θ are positive. The other four functions: cos θ, tan θ, cot θ, and sec θ are negative.
Example #4: Determine the signs of the trigonometric functions of an angle in standard position with the given measure. $$-150°$$ A -150° angle that is in standard position will lie in quadrant III. The x-values and y-values will be negative. Showing a -150 degree angle in standard position Using our chart above, we can see that tan θ and cot θ are positive. The other four functions: sin θ, cos θ, sec θ, and csc θ are negative.

Identifying the Quadrant of an Angle

In some cases, we may be asked to identify the quadrant or possible quadrants of an angle θ that satisfies the given conditions. Let's look at an example.
Example #5: Identify the quadrant(s) of angle θ that satisfies the given conditions.
cos θ > 0, csc θ < 0
If we consult our chart above, we can see that cos θ is positive in quadrants I and IV, however, csc θ is negative in quadrants III and IV. Since quadrant IV meets both conditions, we can state our answer as quadrant IV.

Finding all Trigonometric Function Values Given One Value and the Quadrant

In some cases, we may be asked to find all six trigonometric function values given one function value and the quadrant. Let's look at an example.
Example #6: Use the given information to find the values of all six trigonometric functions.
$$\text{sec}\hspace{.25em}θ=\frac{19}{11}$$ $$\text{sin}\hspace{.25em}θ < 0$$ We know that sin θ is negative in quadrants III and IV. sec θ is given as 19/11, which is positive. sec θ is only positive in quadrants I and IV. Therefore, our angle is in quadrant IV since it meets both conditions. This tells us that our x-values will be positive and our y-values will be negative. Let's think about sec θ $$\text{sec}\hspace{.25em}θ=\frac{r}{x}$$ Let's substitute to find our unknown value for y. $$\text{sec}\hspace{.25em}θ=\frac{r}{x}=\frac{19}{11}$$ $$r=19$$ $$x=11$$ $$r=\sqrt{x^2 + y^2}$$ $$19=\sqrt{11^2 + y^2}$$ Let's square both sides: $$19^2=\left(\sqrt{121 + y^2}\right)^2$$ $$361=121 + y^2$$ $$361 - 121=y^2$$ $$y^2=240$$ $$y=\pm 4\sqrt{15}$$ Since we know that y is negative in quadrant IV, we can state that y is: $$y=-4\sqrt{15}$$ Now that we know the values for x, y, and r, we can find the values for all six trigonometric functions: $$\text{sin}\hspace{.25em}θ=-\frac{4\sqrt{15}}{19}$$ $$\text{cos}\hspace{.25em}θ=\frac{11}{19}$$ $$\text{tan}\hspace{.25em}θ=-\frac{4\sqrt{15}}{11}$$ $$\text{csc}\hspace{.25em}θ=-\frac{19\sqrt{15}}{60}$$ $$\text{cot}\hspace{.25em}θ=-\frac{11\sqrt{15}}{60}$$ $$\text{sec}\hspace{.25em}θ=\frac{19}{11}$$

Skills Check:

Example #1

Find tan θ $$\text{sin}\hspace{.2em}θ=\frac{3}{5}$$ $$\text{cos}\hspace{.2em}θ > 0$$

Please choose the best answer.

A
$$\text{tan}\hspace{.2em}θ=\frac{4}{3}$$
B
$$\text{tan}\hspace{.2em}θ=\frac{3}{4}$$
C
$$\text{tan}\hspace{.2em}θ=-\frac{3}{4}$$
D
$$\text{tan}\hspace{.2em}θ=\frac{6}{7}$$
E
$$\text{tan}\hspace{.2em}θ=-\frac{\sqrt{3}}{7}$$

Example #2

Find cot θ $$\text{sin}\hspace{.2em}θ=-\frac{2\sqrt{5}}{5}$$ $$\text{cos}\hspace{.2em}θ < 0$$

Please choose the best answer.

A
$$\text{cot}\hspace{.2em}θ=\frac{1}{2}$$
B
$$\text{cot}\hspace{.2em}θ=\frac{2}{3}$$
C
$$\text{cot}\hspace{.2em}θ=-2$$
D
$$\text{cot}\hspace{.2em}θ=\frac{\sqrt{5}}{2}$$
E
$$\text{cot}\hspace{.2em}θ=-\frac{\sqrt{5}}{10}$$

Example #3

Find sin θ $$\text{cot}\hspace{.2em}θ=-\frac{4}{3}$$ $$\text{cos}\hspace{.2em}θ < 0$$

Please choose the best answer.

A
$$\text{sin}\hspace{.2em}θ=-4$$
B
$$\text{sin}\hspace{.2em}θ=\frac{3}{4}$$
C
$$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{3}}{4}$$
D
$$\text{sin}\hspace{.2em}θ=\frac{3}{5}$$
E
$$\text{sin}\hspace{.2em}θ=-\frac{5}{4}$$
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