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Common Core: High School - Functions : Trigonometric Functions

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

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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

Example Questions

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

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

$\cos(\theta)=\frac{20}{29}$

find

$\sin(\theta)$ ?

Possible Answers:

$\sin(\theta)=-\frac{21}{20}$

$\sin(\theta)=\frac{29}{21}$

$\sin(\theta)=-\frac{29}{21}$

$\sin(\theta)=-\frac{21}{29}$

$\sin(\theta)=\frac{21}{29}$

Correct answer:

$\sin(\theta)=-\frac{21}{29}$

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

$\cos(\theta)=\frac{20}{29}$

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.

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

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

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,

$\sin(\theta)=-\frac{21}{29}$

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Example Question #6 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{20}{29}$

find

$\tan(\theta)$ ?

Possible Answers:

$\tan(\theta)=-\frac{21}{29}$

$\tan(\theta)=-\frac{20}{21}$

$\tan(\theta)=\frac{21}{20}$

$\tan(\theta)=-\frac{21}{20}$

$\tan(\theta)=\frac{20}{21}$

Correct answer:

$\tan(\theta)=-\frac{21}{20}$

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

$\cos(\theta)=\frac{20}{29}$

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

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

$\sin^2(\theta)+\left(\frac{20}{29}\right)^2=1$

$\sin^2(\theta)+\frac{20^2}{29^2}=1$

$\sin^2(\theta)+\frac{400}{841}=1$

$\sin^2(\theta)+\frac{400}{841}-\frac{400}{841}=1-\frac{400}{841}$

$\sin^2(\theta)=\frac{841-144}{841}$

$\sin^2(\theta)=\frac{441}{841}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{441}{841}}$

$\sin(\theta)=\pm\frac{21}{29}$

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,

$\sin(\theta)=-\frac{21}{29}$

Step 4: Calculate the tangent.

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

$\tan(\theta)=\frac{-\frac{21}{29}}{\frac{20}{29}}$

$\tan(\theta)=-\frac{21}{29}\times\frac{29}{20}$

$\tan(\theta)=-\frac{21}{20}$

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Example Question #6 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

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

find

$\sin(\theta)$ ?

Possible Answers:

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

$\sin(\theta)=-\frac{3}{5}$

$\sin(\theta)=-\frac{5}{4}$

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

$\sin(\theta)=-\frac{5}{3}$

Correct answer:

$\sin(\theta)=-\frac{3}{5}$

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

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

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

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

$\sin^2(\theta)+\left(\frac{4}{5}\right)^2=1$

$\sin^2(\theta)+\frac{4^2}{5^2}=1$

$\sin^2(\theta)+\frac{16}{25}=1$

$\sin^2(\theta)+\frac{16}{25}-\frac{16}{25}=1-\frac{16}{25}$

$\sin^2(\theta)=\frac{25-16}{25}$

$\sin^2(\theta)=\frac{9}{25}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{9}{25}}$

$\sin(\theta)=\pm\frac{3}{5}$

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,

$\sin(\theta)=-\frac{3}{5}$

Report an Error

Example Question #7 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

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

find

$\tan(\theta)$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

$\sin^2(\theta)+\left(\frac{4}{5}\right)^2=1$

$\sin^2(\theta)+\frac{4^2}{5^2}=1$

$\sin^2(\theta)+\frac{16}{25}=1$

$\sin^2(\theta)+\frac{16}{25}-\frac{16}{25}=1-\frac{16}{25}$

$\sin^2(\theta)=\frac{25-16}{25}$

$\sin^2(\theta)=\frac{9}{25}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{9}{25}}$

$\sin(\theta)=\pm\frac{3}{5}$

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,

$\sin(\theta)=-\frac{3}{5}$

Step 4: Calculate the tangent.

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

$\tan(\theta)=\frac{-\frac{3}{5}}{\frac{4}{5}}$

$\tan(\theta)=-\frac{3}{5}\times\frac{5}{4}$

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

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Example Question #11 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{11}{61}$

find

$\sin(\theta)$ ?

Possible Answers:

$\sin(\theta)=\frac{60}{11}$

$\sin(\theta)=-\frac{60}{11}$

$\sin(\theta)=\frac{60}{61}$

$\sin(\theta)=-\frac{61}{60}$

$\sin(\theta)=-\frac{60}{61}$

Correct answer:

$\sin(\theta)=-\frac{60}{61}$

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

$\cos(\theta)=\frac{11}{61}$

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

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

$\sin^2(\theta)+\left(\frac{11}{61}\right)^2=1$

$\sin^2(\theta)+\frac{11^2}{61^2}=1$

$\sin^2(\theta)+\frac{121}{3721}=1$

$\sin^2(\theta)+\frac{121}{3721}-\frac{121}{3721}=1-\frac{121}{3721}$

$\sin^2(\theta)=\frac{3721-121}{3721}$

$\sin^2(\theta)=\frac{3600}{3721}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{3600}{3721}}$

$\sin(\theta)=\pm\frac{60}{61}$

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,

$\sin(\theta)=-\frac{60}{61}$

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Example Question #12 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

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

$\cos(\theta)=\frac{11}{61}$

find

$\tan(\theta)$ ?

Possible Answers:

$\tan(\theta)=\frac{60}{11}$

$\tan(\theta)=\frac{11}{60}$

$\tan(\theta)=-\frac{61}{11}$

$\tan(\theta)=-\frac{11}{60}$

$\tan(\theta)=-\frac{60}{11}$

Correct answer:

$\tan(\theta)=-\frac{60}{11}$

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

$\cos(\theta)=\frac{11}{61}$

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

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

$\sin^2(\theta)+\left(\frac{11}{61}\right)^2=1$

$\sin^2(\theta)+\frac{11^2}{61^2}=1$

$\sin^2(\theta)+\frac{121}{3721}=1$

$\sin^2(\theta)+\frac{121}{3721}-\frac{121}{3721}=1-\frac{121}{3721}$

$\sin^2(\theta)=\frac{3721-121}{3721}$

$\sin^2(\theta)=\frac{3600}{3721}$

$\sqrt{\sin^2(\theta)}=\sqrt{\frac{3600}{3721}}$

$\sin(\theta)=\pm\frac{60}{61}$

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,

$\sin(\theta)=-\frac{60}{61}$

Step 4: Calculate the tangent.

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

$\tan(\theta)=\frac{-\frac{60}{61}}{\frac{11}{61}}$

$\tan(\theta)=-\frac{60}{61}\times\frac{61}{11}$

$\tan(\theta)=-\frac{60}{11}$

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Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

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

$\sin(105^\circ)$ .

Possible Answers:

$\sin(105^\circ)=\frac{2+\sqrt{6}}{2}$

$\sin(105^\circ)=\frac{\sqrt{8}}{2}$

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{2}$

$\sin(105^\circ)=\frac{\sqrt{2}}{2}$

Correct answer:

$\sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

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.

$105^\circ=60^\circ+45^\circ$

Step 2: Write the general addition formula for sine.

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

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

$\\a=60^\circ \\b=45^\circ$

$\\ \sin(60^\circ+45^\circ)=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ) \\\sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{2}\times\frac{1}{\sqrt{2}}+\frac{1}{2}\times\frac{1}{\sqrt{2}} \\ \\ \sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ \\ \sin(60^\circ+45^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}$

To rationalize the denominator multiply the numerator and denominator by the square root of two.

$\\ \sin(60^\circ+45^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4} \\ \\ \sin(105^\circ)=\frac{\sqrt{2}+\sqrt{6}}{4}$

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Example Question #2 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

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

$\sin(75^\circ)$ .

Possible Answers:

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2\sqrt2}$

$\sin(75^\circ)=\frac{1+\sqrt{3}}{\sqrt2}$

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2}$

$\sin(75^\circ)=\frac{1}{2\sqrt2}$

$\sin(75^\circ)=\frac{\sqrt{3}}{2\sqrt2}$

Correct answer:

$\sin(75^\circ)=\frac{1+\sqrt{3}}{2\sqrt2}$

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.

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

$75^\circ=30^\circ+45^\circ$

Step 2: Write the general addition formula for sine.

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

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

$\\a=30^\circ \\b=45^\circ$

$\\ \sin(30^\circ+45^\circ)=\sin(30^\circ)\cos(45^\circ)+\cos(30^\circ)\sin(45^\circ) \\\sin(30^\circ+45^\circ)=\frac{1}{2}\times\frac{1}{\sqrt{2}}+\frac{\sqrt3}{2}\times\frac{1}{\sqrt{2}} \\ \\ \sin(30^\circ+45^\circ)=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}} \\ \\ \sin(30^\circ+45^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}$

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Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

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

$\sin(120^\circ)$ .

Possible Answers:

$\sin(120^\circ)=\frac{\sqrt{3}}{2\sqrt{2}}$

$\sin(120^\circ)=\frac{\sqrt{3}}{2}$

$\sin(120^\circ)=-\frac{\sqrt{3}}{4}$

$\sin(120^\circ)=-\frac{\sqrt{3}}{2}$

$\sin(120^\circ)=\frac{\sqrt{3}}{4}$

Correct answer:

$\sin(120^\circ)=\frac{\sqrt{3}}{2}$

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.

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

$120^\circ=60^\circ+60^\circ$

Step 2: Write the general addition formula for sine.

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

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

$\\a=60^\circ \\b=60^\circ$

$\\ \sin(60^\circ+60^\circ)=\sin(60^\circ)\cos(60^\circ)+\cos(60^\circ)\sin(60^\circ) \\\sin(60^\circ+60^\circ)=\frac{\sqrt{3}}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2} \\ \\ \sin(60^\circ+45^\circ)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \\ \sin(60^\circ+60^\circ)=\frac{2\sqrt{3}}{4}$

$\\ \sin(60^\circ+60^\circ)=\frac{\sqrt{3}}{2}\\ \\ \sin(120^\circ)=\frac{\sqrt{3}}{2}$

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Example Question #4 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

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

$\sin(135^\circ)$ .

Possible Answers:

$\sin(135^\circ)=-\frac{1}{2}$

$\sin(135^\circ)=-\frac{1}{\sqrt{2}}$

$\sin(135^\circ)=\frac{1}{2}$

$\sin(135^\circ)={\sqrt{2}}$

$\sin(135^\circ)=\frac{1}{\sqrt{2}}$

Correct answer:

$\sin(135^\circ)=\frac{1}{\sqrt{2}}$

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.

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 sine.

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

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

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

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

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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 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #6 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #6 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #7 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #11 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #12 : Prove Pythagorean Identity: Ccss.Math.Content.Hsf Tf.C.8

Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #2 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #1 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

Example Question #4 : Prove Addition And Subtraction Formula For Trigonometric Functions: Ccss.Math.Content.Hsf Tf.C.9

All Common Core: High School - Functions Resources

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