Trigonometric Functions

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Common Core: High School - Functions › Trigonometric Functions

Questions 1 - 10

If $\theta$ lives in quadrant IV and cosine is as follows,

$\cos(\theta)=\frac{12}{13}$

find

$\tan(\theta)$ ?

$\tan(\theta)=-\frac{5}{12}$

$\tan(\theta)=\frac{5}{12}$

$\tan(\theta)=-\frac{5}{13}$

$\tan(\theta)=\frac{5}{13}$

$\tan(\theta)=-\frac{12}{5}$

Explanation

This question tests one's understand of the Pythagorean Theorem as it relates to trigonometric functions. It also uses the method of algebraic manipulation of functions with the aid of trigonometric identities to simplify and solve the problem in question. For questions like these, it is important to recall that the Pythagorean Theorem, as it relates to trigonometric functions is

$\sin^2(\theta)+\cos^2(\theta)=1$

and holds true for any value of $\theta$ .

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For the purpose of Common Core Standards, "prove the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ and use it to fine sine, cosine, or tangent and the quadrant of the angle", falls within the Cluster C of "prove and apply trigonometric identities" concept (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify given information.

Given

$\cos(\theta)=\frac{12}{13}$

The question states the angle is in quadrant IV which means sine and tangent are negative while cosine is positive. (An easy way to remember this is to say "All Students Take Calculus"; this means that in quadrant I all trigonometric functions are positive, in quadrant II only sine and its inverse are positive, in quadrant III only tangent and its inverse are positive, and in quadrant IV only cosine and its inverse are positive. ASTC)

Step 2: Plug in given information into Pythagorean identity and use algebraic manipulation to solve for the unknown.

$\sin^2(\theta)+\cos^2(\theta)=1$

$\sin^2(\theta)+\left(\frac{12}{13}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{13^2}=1$

$\sin^2(\theta)+\frac{144}{169}=1$

$\sin^2(\theta)+\frac{144}{169}-\frac{144}{169}=1-\frac{144}{169}$

$\sin^2(\theta)=\frac{169-144}{169}$

$\sin^2(\theta)=\frac{25}{169}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{25}{169}}$

$\sin(\theta)=\pm\frac{5}{13}$

Step 3: Using the quadrant information, solve the question.

Since the question states that the angle is in quadrant IV, that means that sine will be negative.

Therefore,

$\sin(\theta)=-\frac{5}{13}$

Step 4: Calculate the tangent.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{-\frac{5}{13}}{\frac{12}{13}}$

$\tan(\theta)=-\frac{5}{13}\times\frac{13}{12}$

$\tan(\theta)=-\frac{5}{12}$

Calculate sine of the following angle.

$\theta=\frac{3\pi}{9}$

$\sin\left(\frac{3\pi}{9} \right )=\frac{\sqrt{3}}{2}$

$\sin\left(\frac{3\pi}{9} \right )=-\frac{\sqrt{3}}{2}$

$\sin\left(\frac{3\pi}{9} \right )=\frac{1}{2}$

$\sin\left(\frac{3\pi}{9} \right )=\frac{-1}{2}$

$\sin\left(\frac{3\pi}{9} \right )=\sqrt{3}$

Explanation

This question tests one's ability to understand the connection between special triangles (30-60-90 and 45-45-90), trigonometric functions related to them (sine, cosine, tangent), and the corresponding degree/radian on the unit circle. It relies on the understanding that the measure of an angle can be described using the properties of the unit circle and arc length. Furthermore, questions involving concepts dealing with trigonometry use the foundation of corresponding identities with angles that come from the unit circle. Recall that the unit circle is a circle that has a radius of one and is centered at the origin. Each pair that lies on the circle can be found by creating a right triangle that has a base on the -axis and a hypotenuse that is from the center of the circle to the desired point on the edge of the circle. The height of the triangle is the vertical line that connects the point on the circle's edge to the -axis.

For the purpose of Common Core Standards, understanding "use special triangles to determine geometrically the values of sine, cosine, tangent for $\frac{\pi}{3}, \frac{\pi}{4},\frac{\pi}{6}$ " , falls within the Cluster A of "extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF.TF.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Possible approaches to solving this problem:

I. Plot the angle on the unit circle to find the $(x,y)=(cos(\theta),sin(\theta))$ .

II. Identify the special triangle that corresponds to the angle.

III. Using memorization of the unit circle.

For this particular question let's us approach II.

Step 1: Simplify the angle if possible.

$\theta=\frac{3\pi}{9}=\frac{3\cdot 1\pi}{3\cdot 3}=\frac{\pi}{3}$

Step 2: Identify the special right triangle that corresponds to the angle.

Let's convert the angle to degrees to more easily see the special right triangle that it creates.

$\ \frac{\pi}{3}\times \frac{180^\circ}{\pi}\ \=\frac{3\cdot 60}{3}\\=60^\circ$

Therefore the special triangle that is created with $\theta$ is 30-60-90 degree triangle. The angle which is made at the origin will be 60 degrees and the length of the segment from the origin to the edge of the unit circle is one by definition.

Step 3: Calculate the side lengths of the special triangle.

From here, recall that the special 30-60-90 degree triangle has the following side identities.

$\\textup{short side}= \frac{1}{2}\times\textup{hypotenuse} \\textup{long side}=\textup{short side}\times \sqrt{3}$

Substituting in one for the hypotenuse, the short and long leg of the triangle can be found.

$\\textup{short side}=\frac{1}{2}\times 1=\frac{1}{2} \ \ \textup{long side}=\frac{1}{2}\times\sqrt{3}=\frac{\sqrt{3}}{2}$

It is important to remember that the short side of the triangle is on the -axis and adjacent to $\theta$ . The long side is the vertical line that is parallel to the -axis and opposite of $\theta$ .

Step 4: Calculate sine, cosine, and tangent.

Recall the following trigonometric identities.

$\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}$

From Step 2,

$\ \textup{opposite}=\frac{\sqrt{3}}{2} \ \ \textup{adjacent}=\frac{1}{2} \ \ \textup{hypotenuse}=1$

Substituting the values found in Step 2 into the above identities is as follows.

$\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$

Calculate cosine of the following angle.

$\theta=\frac{6\pi}{8}$

$\cos\left(\frac{6\pi}{8} \right )=-\frac{1}{\sqrt{2}}$

$\cos\left(\frac{6\pi}{8} \right )=\frac{1}{\sqrt{2}}$

$\cos\left(\frac{6\pi}{8} \right )=-\frac{1}{2}$

$\cos\left(\frac{6\pi}{8} \right )=\frac{1}{2}$

$\cos\left(\frac{6\pi}{8} \right )=-1$

Explanation

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Possible approaches to solving this problem:

I. Plot the angle on the unit circle to find the $(x,y)=(cos(\theta),sin(\theta))$ .

II. Identify the special triangle that corresponds to the angle.

III. Using memorization of the unit circle.

For this particular question let's us approach II.

Step 1: Simplify the angle if possible.

$\theta=\frac{6\pi}{8}=\frac{2\cdot 3\pi}{2\cdot 4}=\frac{3\pi}{4}$

Step 2: Identify the special right triangle that corresponds to the angle.

Let's convert the angle to degrees to more easily see the special right triangle that it creates.

$\ \frac{3\pi}{4}\times \frac{180^\circ}{\pi}\ \=\frac{4\cdot 135}{4}\\=135^\circ$

Find the reference angle by subtracting the angle from from 180 degrees.

$180^\circ-135^\circ=45^\circ$

This means that 135 degrees is a 45 degree angle that lies in the second quadrant. Recall that the second quadrant contains positive values and negative values.

Therefore the special triangle that is created with $\theta$ is 45-45-90 degree triangle. The angle which is made at the origin will be 45 degrees and the length of the segment from the origin to the edge of the unit circle is one by definition.

Step 3: Calculate the side lengths of the special triangle.

From here, recall that the special 45-45-90 degree triangle has the following side identities.

$\ \text{side lengths}: \\text{opposite side}=\text{adjacent side}=1 \ \text{hypotenuse}=\sqrt{2}$

Step 4: Calculate cosine.

Recall the following trigonometric identities.

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}$

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{1}{\sqrt{2}}$

Since cosine represents the values on the coordinate grid, this means that cosine is negative for angles in quadrant two.

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=-\frac{1}{\sqrt{2}}$

Using the addition formula for cosine and special reference angles calculate,

$\cos(135^\circ)$ .

$\cos(135)=-\frac{1}{\sqrt{2}}$

$\cos(135)=\frac{1}{\sqrt{2}}$

$\cos(135)=-\frac{1}{2}$

$\cos(135)=\frac{1}{2}$

$\cos(135)=-\frac{1}{2\sqrt{2}}$

Explanation

This type of question tests the deep understanding of geometry, right triangles, trigonometry, and dealing with proofs. Questions such as these are not designed to be tested but instead are used to build knowledge that will help in higher level mathematics courses.

For the purpose of Common Core Standards, "prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems", falls within the Cluster C of "prove and apply trigonometric identities" (CCSS.MATH.CONTENT.HSF.TF.C).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Break the angle into two angles that correspond to special reference angles.

$135^\circ=90^\circ+45^\circ$

Step 2: Write the general addition formula for cosine.

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Step 3: Substitute in the reference angles into the general addition formula for cosine.

$\a=90^\circ \b=45^\circ$

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(90+45)=\cos(90)\cos(45)-\sin(90)\sin(45)$

$\cos(90+45)=0\times \frac{1}{\sqrt{2}}-1\times \frac{1}{\sqrt{2}}$

$\cos(90+45)=-\frac{1}{\sqrt{2}}$

$\cos(135)=-\frac{1}{\sqrt{2}}$

$\theta$ is the angle between the -axis and the line that connects the origin to the point .

Calculate $\tan(\theta)$ .

$\tan(\theta)=\frac{4}{3}$

$\tan(\theta)=\frac{3}{4}$

$\tan(\theta)=\frac{4}{5}$

$\tan(\theta)=\frac{3}{5}$

$\tan(\theta)=\frac{5}{3}$

Explanation

This question is testing one's ability to understand the trigonometric relationships and how the relate to the unit circle for solving problems.

For the purpose of Common Core Standards, "Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.2).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Plot the line that connects the origin to the point .

Screen shot 2016 01 14 at 11.03.26 am

Step 2: Use right triangles and the unit circle to identify the trigonometric characteristics of the describe situation.

The unit circle is an extremely helpful tool in solving trigonometric problems. For the purpose of trigonometry, the unit circle is located at the origin and has a radius of one unit. This is because right triangles can be formed using the x axis as one leg of the triangle and the line from the origin to a point on the circle as the hypotenuse with a measurement of one.

Therefore, the point on the circle has coordinates of $(\cos(x),\sin(x))$ for the angle that is created by the x axis and the hypotenuse. It is important to know that

$(x,y)=(\cos(\theta),\sin(\theta))$ and can be found using the triangle that is created on the unit circle.

Recall that using trigonometric identities tangent is,

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

For this particular question,

$\(3,4)=(\cos(\theta),\sin(\theta)) \\cos(\theta)=3 \\sin(\theta)=4$

Step 3: Answer the question.

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

$\tan(\theta)=\frac{4}{3}$

Find the exact value of the following statement.

$f(x)=\sin^{-1}(-1)$

$\sin^{-1}(-1)=\frac{3\pi}{2}$

$\sin^{-1}(-1)=\frac{\pi}{2}$

$\sin^{-1}(-1)=\frac{3\pi}{4}$

$\sin^{-1}(-1)=\pi$

$\sin^{-1}(-1)=2\pi$

Explanation

This question is testing ones ability to understand and identify inverses of trigonometric functions as they relate to the unit circle.

For the purpose of Common Core Standards, " Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed." concept (CCSS.MATH.CONTENT.HSF-TF.B.6). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify what the question is asking for.

Since there is a trigonometric function raised to the negative one power, this question is talking about the inverse of the function. In other words, which angle on the unit circle results in a sine equalling negative one?

$y=\sin^{-1}(a)\Rightarrow \sin(\theta)=a$

Therefore, theta needs to be solved for.

Step 2: Draw and label the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 3: Locate the angle that results in negative one for its sine value.

Recall that

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore look for the that has . Looking at the unit circle from step 2, it is seen that at angle $\frac{3\pi}{2}$ the sine equals negative one.

Thus,

$f(x)=\sin^{-1}(-1)\Rightarrow f(x)=\frac{3\pi}{2}$

To verify the solution simply find the sine of the angle theta.

$\sin\left(\frac{3\pi}{2}\right)=-1\rightarrow\sin^{-1}(-1)=\frac{3\pi}{2}$

Solve for $\theta$ .

$\tan(\theta)+1=-\sqrt{3}+1$

$\theta=\frac{2\pi}{3},\frac{5\pi}{3}$

$\theta=\frac{2\pi}{3},\frac{5\pi}{6}$

$\theta=\frac{2\pi}{6},\frac{5\pi}{3}$

$\theta=\frac{2\pi}{6},\frac{5\pi}{6}$

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

Explanation

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)+1=-\sqrt{3}+1$

Subtract one from each side.

$\tan(\theta)+1{\color{Red} -1}=-\sqrt{3}+1{\color{Red} -1}$

$\tan(\theta)=-\sqrt{3}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ \tan^{-1}(\tan(\theta))=\tan^{-1}(-\sqrt{3})$

$\theta=\tan^{-1}(-\sqrt{3})$

Step 3: Use the unit circle to solve for theta.

Screen shot 2016 01 14 at 10.52.42 am

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left(\frac{2\pi}{3} \right )=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\frac{\sqrt{3}}{2}\cdot -\frac{2}{1}=-\sqrt{3}$

$\tan\left(\frac{5\pi}{3} \right )=\frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{-\sqrt{3}}{2}\cdot \frac{2}{1}=-\sqrt{3}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{2\pi}{3},\frac{5\pi}{3}$

Find the following trigonometric exact value.

$\cos\left(\frac{5\pi}{6} \right )$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

$\cos\left(\frac{5\pi}{6} \right )=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{1}{2}$

$\cos\left(\frac{5\pi}{6} \right )=\frac{1}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\sqrt{3}$

Explanation

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the coordinate pair that represents the extension of $\theta = \frac{5\pi}{6}$ .

$\theta = \frac{5\pi}{6}\Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{2} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{5\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)=\left(\cos\left(\frac{5\pi}{6}\right),\sin\left(\frac{5\pi}{6}\right)\right)$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$ .

Calculate cosine of the following angle.

$\theta=\frac{6\pi}{8}$

$\cos\left(\frac{6\pi}{8} \right )=-\frac{1}{\sqrt{2}}$

$\cos\left(\frac{6\pi}{8} \right )=\frac{1}{\sqrt{2}}$

$\cos\left(\frac{6\pi}{8} \right )=-\frac{1}{2}$

$\cos\left(\frac{6\pi}{8} \right )=\frac{1}{2}$

$\cos\left(\frac{6\pi}{8} \right )=-1$

Explanation

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Possible approaches to solving this problem:

I. Plot the angle on the unit circle to find the $(x,y)=(cos(\theta),sin(\theta))$ .

II. Identify the special triangle that corresponds to the angle.

III. Using memorization of the unit circle.

For this particular question let's us approach II.

Step 1: Simplify the angle if possible.

$\theta=\frac{6\pi}{8}=\frac{2\cdot 3\pi}{2\cdot 4}=\frac{3\pi}{4}$

Step 2: Identify the special right triangle that corresponds to the angle.

Let's convert the angle to degrees to more easily see the special right triangle that it creates.

$\ \frac{3\pi}{4}\times \frac{180^\circ}{\pi}\ \=\frac{4\cdot 135}{4}\\=135^\circ$

Find the reference angle by subtracting the angle from from 180 degrees.

$180^\circ-135^\circ=45^\circ$

This means that 135 degrees is a 45 degree angle that lies in the second quadrant. Recall that the second quadrant contains positive values and negative values.

Step 3: Calculate the side lengths of the special triangle.

From here, recall that the special 45-45-90 degree triangle has the following side identities.

$\ \text{side lengths}: \\text{opposite side}=\text{adjacent side}=1 \ \text{hypotenuse}=\sqrt{2}$

Step 4: Calculate cosine.

Recall the following trigonometric identities.

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}$

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{1}{\sqrt{2}}$

Since cosine represents the values on the coordinate grid, this means that cosine is negative for angles in quadrant two.

$\ \cos(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=-\frac{1}{\sqrt{2}}$

Find the following trigonometric exact value.

$\cos\left(\frac{5\pi}{6} \right )$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$

$\cos\left(\frac{5\pi}{6} \right )=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{1}{2}$

$\cos\left(\frac{5\pi}{6} \right )=\frac{1}{2}$

$\cos\left(\frac{5\pi}{6} \right )=-\sqrt{3}$

Explanation

This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Recall the unit circle.

Screen shot 2016 01 14 at 10.52.42 am

Step 2: Identify the coordinate pair that represents the extension of $\theta = \frac{5\pi}{6}$ .

$\theta = \frac{5\pi}{6}\Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{2} \right )$

Step 3: Calculate the exact value of $\cos\left(\frac{5\pi}{6} \right )$ .

Recall that,

$(x,y)=(\cos(\theta),\sin(\theta))$

therefore,

$\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)=\left(\cos\left(\frac{5\pi}{6}\right),\sin\left(\frac{5\pi}{6}\right)\right)$

$\cos\left(\frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}$ .

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