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All Common Core: High School - Functions Resources

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Flashcards: 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,

$\sin(105^\circ)$ .

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

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

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

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

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

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Flashcards: 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,

$\sin(105^\circ)$ .

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

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.

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

Flashcards: 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,

$\sin(75^\circ)$ .

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

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

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

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

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

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Flashcards: 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,

$\sin(75^\circ)$ .

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

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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All Common Core: High School - Functions Resources

Flashcards: Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

Flashcards: Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

Flashcards: Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

Flashcards: Prove Addition and Subtraction Formula for Trigonometric Functions: CCSS.Math.Content.HSF-TF.C.9

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