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Common Core: High School - Functions : Trigonometric Functions

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Example Questions

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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,

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

Report an Error

Example Question #3 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

$2\tan(\theta)=-2\sqrt{3}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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.

$2\tan(\theta)=-2\sqrt{3}$

Divide by two on each side.

$\frac{2\tan(\theta)}{2}=\frac{-2\sqrt{3}}{2}$

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

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,

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

Report an Error

Example Question #11 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

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

Possible Answers:

$\theta=\frac{5\pi}{3},\frac{11\pi}{3}$

$\theta=\frac{5\pi}{6},\frac{7\pi}{6}$

$\theta=\frac{5\pi}{3},\frac{11\pi}{6}$

$\theta=\frac{5\pi}{6},\frac{11\pi}{6}$

$\theta=\frac{5\pi}{6},\frac{11\pi}{3}$

Correct answer:

$\theta=\frac{5\pi}{6},\frac{11\pi}{6}$

Explanation:

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

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.

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

Subtract one from each side.

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

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

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

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

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

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,

$\tan\left(\frac{5\pi}{6} \right )=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{1}{2}\cdot -\frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

$\tan\left(\frac{11\pi}{6} \right )=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$

Step 4: Answer the question.

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

$\theta=\frac{5\pi}{6},\frac{11\pi}{6}$

Report an Error

Example Question #12 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Solve for $\theta$ .

$-\tan(\theta)+2=3$

Possible Answers:

$\theta=\frac{\pi}{4},\frac{7\pi}{4}$

$\theta=\frac{3\pi}{2},\frac{7\pi}{2}$

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

$\theta=\frac{3\pi}{4},\frac{7\pi}{3}$

$\theta=\frac{3\pi}{2},\frac{7\pi}{4}$

Correct answer:

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Explanation:

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

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.

$-\tan(\theta)+2=3$

Subtract two from both sides.

$-\tan(\theta)+2{\color{Red} -2}=3{\color{Red} -2}$

$-\tan(\theta)=1$

Now divide by negative one on both sides.

$\tan(\theta)=-1$

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

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

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

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,

$\tan\left(\frac{3\pi}{4} \right )=\frac{\frac{1}{\sqrt2}}{-\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}\cdot -\frac{\sqrt2}{1}=-1$

$\tan\left(\frac{7\pi}{4} \right )=\frac{-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=-\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{1}=-1$ .

Step 4: Answer the question.

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

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Report an Error

Example Question #1 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

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

find

$\sin(\theta)$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\theta$ .

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,

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

Report an Error

Example Question #81 : Trigonometric Functions

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

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

find

$\tan(\theta)$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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.

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

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

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

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

Report an Error

Example Question #2 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{8}{17}$

find

$\sin(\theta)$ ?

Possible Answers:

$\sin(\theta)=-\frac{17}{15}$

$\sin(\theta)=\frac{17}{15}$

$\sin(\theta)=-\frac{15}{17}$

$\sin(\theta)=\frac{15}{17}$

$\sin(\theta)=-\frac{8}{15}$

Correct answer:

$\sin(\theta)=-\frac{15}{17}$

Explanation:

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

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

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

$\cos(\theta)=\frac{8}{17}$

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

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

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

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,

$\sin(\theta)=-\frac{15}{17}$

Report an Error

Example Question #1 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{8}{17}$

find

$\tan(\theta)$

Possible Answers:

$\tan(\theta)=\frac{15}{8}$

$\tan(\theta)=-\frac{15}{8}$

$\tan(\theta)=-\frac{15}{17}$

$\tan(\theta)=\frac{8}{15}$

$\tan(\theta)=\frac{15}{17}$

Correct answer:

$\tan(\theta)=-\frac{15}{8}$

Explanation:

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

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

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

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

$\cos(\theta)=\frac{8}{17}$

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

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

$\sin^2(\theta)+\left(\frac{8}{17}\right)^2=1$

$\sin^2(\theta)+\frac{8^2}{17^2}=1$

$\sin^2(\theta)+\frac{64}{289}=1$

$\sin^2(\theta)+\frac{64}{289}-\frac{64}{289}=1-\frac{64}{289}$

$\sin^2(\theta)=\frac{289-64}{289}$

$\sin^2(\theta)=\frac{225}{289}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{225}{289}}$

$\sin(\theta)=\pm\frac{15}{17}$

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,

$\sin(\theta)=-\frac{15}{17}$

Step 4: Calculate the tangent.

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

$\tan(\theta)=\frac{-\frac{15}{17}}{\frac{8}{17}}$

$\tan(\theta)=-\frac{15}{17}\times\frac{17}{8}$

$\tan(\theta)=-\frac{15}{8}$

Report an Error

Example Question #4 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{12}{37}$

find

$\sin(\theta)$ ?

Possible Answers:

$\sin(\theta)=-\frac{35}{12}$

$\sin(\theta)=\frac{35}{37}$

$\sin(\theta)=-\frac{37}{35}$

$\sin(\theta)=\frac{37}{35}$

$\sin(\theta)=-\frac{35}{37}$

Correct answer:

$\sin(\theta)=-\frac{35}{37}$

Explanation:

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

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

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

$\cos(\theta)=\frac{12}{37}$

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

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

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

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,

$\sin(\theta)=-\frac{35}{37}$

Report an Error

Example Question #5 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{12}{37}$

find

$\tan(\theta)$ ?

Possible Answers:

$\tan(\theta)=-\frac{12}{35}$

$\tan(\theta)=-\frac{35}{12}$

$\tan(\theta)=\frac{12}{35}$

$\tan(\theta)=\frac{35}{12}$

$\tan(\theta)=-\frac{35}{37}$

Correct answer:

$\tan(\theta)=-\frac{35}{12}$

Explanation:

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

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

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

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

$\cos(\theta)=\frac{12}{37}$

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

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

$\sin^2(\theta)+\left(\frac{12}{37}\right)^2=1$

$\sin^2(\theta)+\frac{12^2}{37^2}=1$

$\sin^2(\theta)+\frac{144}{1369}=1$

$\sin^2(\theta)+\frac{144}{1369}-\frac{144}{1369}=1-\frac{144}{1369}$

$\sin^2(\theta)=\frac{1369-144}{1369}$

$\sin^2(\theta)=\frac{1225}{1369}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{1225}{1369}}$

$\sin(\theta)=\pm\frac{35}{37}$

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,

$\sin(\theta)=-\frac{35}{37}$

Step 4: Calculate the tangent.

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

$\tan(\theta)=\frac{-\frac{35}{37}}{\frac{12}{37}}$

$\tan(\theta)=-\frac{35}{37}\times\frac{37}{12}$

$\tan(\theta)=-\frac{35}{12}$

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All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

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Common Core: High School - Functions : Trigonometric Functions

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #1 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #3 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #11 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #12 : Solve Trigonometric Functions With Inverse Functions: Ccss.Math.Content.Hsf Tf.B.7

Example Question #1 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #81 : Trigonometric Functions

Example Question #2 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #1 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #4 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #5 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

All Common Core: High School - Functions Resources

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