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Common Core: High School - Functions : Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

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Example Questions

Example Question #461 : High School: Functions

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

$\cos(120^\circ)$ .

Possible Answers:

$\cos(120^\circ)=\frac{1}{\sqrt2}$

$\cos(120^\circ)=-1$

$\cos(120^\circ)=-\frac{1}{\sqrt{2}}$

$\cos(120^\circ)=\frac{1}{2}$

$\cos(120^\circ)=-\frac{1}{2}$

Correct answer:

$\cos(120^\circ)=-\frac{1}{2}$

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.

$120^\circ=90^\circ+30^\circ$

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.

$\\a=90^\circ \\b=30^\circ$

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

$\cos(90+30)=\cos(90)\cos(30)-\sin(90)\sin(30)$

$\cos(90+30)=0\times\frac{\sqrt{3}}{2}-1\times\frac{1}{2}$

$\cos(90+30)=0-\frac{1}{2}$

$\cos(90+30)=-\frac{1}{2}$

$\cos(120^\circ)=-\frac{1}{2}$

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Example Question #462 : High School: Functions

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

$\cos(135^\circ)$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 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.

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

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

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

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

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

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

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Common Core: High School - Functions : Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

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

All Common Core: High School - Functions Resources

Example Questions

Example Question #461 : High School: Functions

Example Question #462 : High School: Functions

All Common Core: High School - Functions Resources

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