Trigonometric Functions - Common Core: High School - Functions

Card 0 of 20

Question

What is the period of the following function?

$y=\sin(2x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the period.

Given the function

$y=\sin(2x)$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=2\rightarrow \frac{2\pi}{|2|}=\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the period of this function is $\pi$ because the flow of the graph only begins to repeat its cycle after $\pi$ units on the - axis.

Compare your answer with the correct one above

Question

What is the amplitude of the following function?

$y=2\sin(x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the amplitude.

Given the function

$y=2\sin(x)$

identify the variables of the general form.

$\f(x)=\sin \a=2 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the amplitude of this function is two because the range of the function on the graph goes from negative two to positive two meaning the distance from zero at its highest peak or lowest valley is two.

Compare your answer with the correct one above

Question

What is the amplitude of the following function?

$y=4\sin(x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the amplitude.

Given the function

$y=4\sin(x)$

identify the variables of the general form.

$\f(x)=\sin \a=4 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the amplitude of this function is four because the range of the function on the graph goes from negative four to positive four meaning the distance from zero at its highest peak or lowest valley is four.

Compare your answer with the correct one above

Question

What is the period of the following function?

$y=\sin(4x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the period.

Given the function

$y=\sin(4x)$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=4\rightarrow \frac{2\pi}{|4|}=\frac{\pi}{2} \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the period of this function is $\frac{\pi}{2}$ because the flow of the graph only begins to repeat its cycle after $\frac{\pi}{2}$ units on the - axis.

Compare your answer with the correct one above

Question

What is the vertical shift of the function?

$f(x)=\sin(x)+\pi$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the vertical shift.

Given the function

$y=\sin(x)+\pi$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=\pi$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $\pi$ .

Compare your answer with the correct one above

Question

What is the vertical shift of the function?

$f(x)=\sin(x)-\pi$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the vertical shift.

Given the function

$y=\sin(x)-\pi$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=-\pi$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $-\pi$ .

Compare your answer with the correct one above

Question

What is the horizontal shift of the function?

$f(x)=\sin(x-\pi)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the horizontal shift.

Given the function

$y=\sin(x-\pi)$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=\pi \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the horizontal shift of the function is $\pi$ .

Compare your answer with the correct one above

Question

What is the horizontal shift of the function?

$f(x)=\sin(x+\pi)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the horizontal shift.

Given the function

$y=\sin(x+\pi)$

identify the variables of the general form.

$\f(x)=\sin \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=-\pi \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the horizontal shift of the function is $-\pi$ .

Compare your answer with the correct one above

Question

What is the amplitude of the following function?

$y=2\cos(x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the amplitude.

Given the function

$y=2\cos(x)$

identify the variables of the general form.

$\f(x)=\cos \a=2 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the amplitude of this function is two because the range of the function on the graph goes from negative two to positive two meaning the distance from zero at its highest peak or lowest valley is two.

Compare your answer with the correct one above

Question

What is the period of the following function?

$y=\cos(2x)$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the period.

Given the function

$y=\cos(2x)$

identify the variables of the general form.

$\f(x)=\cos \a=1 \b=2\rightarrow \frac{2\pi}{|2|}=\pi \c=0 \d=0$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the period of this function is $\pi$ because the flow of the graph only begins to repeat its cycle after $\pi$ units on the - axis.

Compare your answer with the correct one above

Question

What is the vertical shift of the function?

$f(x)=\cos(x)+\pi$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the vertical shift.

Given the function

$y=\cos(x)+\pi$

identify the variables of the general form.

$\f(x)=\cos \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=\pi$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $\pi$ .

Compare your answer with the correct one above

Question

What is the vertical shift of the function?

$f(x)=\cos(x)-\pi$

Answer

This question is testing one's ability to identify the periodicity of a trigonometric function. This requires the understanding of trigonometric functions and their graphical and algebraic characteristics.

For the purpose of Common Core Standards, "Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline" falls within the Cluster B of "Model Periodic Phenomena with Trigonometric Functions" concept (CCSS.MATH.CONTENT.HSF-TF.B.5).

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: Write the general form of trigonometric shifts.

where

$\F(x)=\textup{New function after shift} \a=\textup{Amplitude} \b=\textup{period}\rightarrow \frac{2\pi}{|B|} \ \c=\textup{Horizontal shift} \ d=\textup{Vertical shift} \f(x)=\textup{Trigonometric function}$

Step 2: Algebraically identify the vertical shift.

Given the function

$y=\cos(x)-\pi$

identify the variables of the general form.

$\f(x)=\cos \a=1 \b=1\rightarrow \frac{2\pi}{|1|}=2\pi \c=0 \d=-\pi$

Step 3: Graph the trigonometric function to verify.

The graph above verifies that the vertical shift of the function is $-\pi$ .

Compare your answer with the correct one above

Question

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

$\sin(105^\circ)$ .

Answer

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.

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

Compare your answer with the correct one above

Question

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

$\sin(75^\circ)$ .

Answer

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.

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

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.

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

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

Compare your answer with the correct one above

Question

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

$\sin(120^\circ)$ .

Answer

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.

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

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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Question

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

$\sin(135^\circ)$ .

Answer

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 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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Question

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

$\sin(450^\circ)$ .

Answer

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.

$450^\circ=360^\circ+90^\circ$

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.

$\a=360^\circ \b=90^\circ$

$\sin(360+90)=\sin(360)\cos(90)+\cos(360)\sin(90)$

$\sin(360+90)=0\times 0+1\times1$

$\sin(360+90)=1$

$\sin(450^\circ)=1$

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Question

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

$\sin(15^\circ)$ .

Answer

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.

$15^\circ=45^\circ-30^\circ$

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.

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

$\sin(45-30)=\sin(45)\cos(30)-\cos(45)\sin(30)$

$\sin(45-30)=\frac{1}{\sqrt{2}}\times \frac{\sqrt3}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$

$\sin(45-30)= \frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}$

$\sin(45-30)= \frac{\sqrt3-1}{2\sqrt{2}}$

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

$\sin(45-30)= \frac{\sqrt3-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$\sin(45-30)= \frac{\sqrt6-\sqrt{2}}{2\times 2}$

$\sin(45-30)= \frac{\sqrt6-\sqrt{2}}{4}$

$\sin(15^\circ)= \frac{\sqrt6-\sqrt{2}}{4}$

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Question

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

$\cos(15^\circ)$ .

Answer

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.

$15^\circ=45^\circ-30^\circ$

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.

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

$\cos(45-30)=\cos(45)\cos(30)+\sin(45)\sin(30)$

$\cos(45-30)=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}$

$\cos(45-30)=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}$

$\cos(45-30)=\frac{\sqrt{3}+1}{2\sqrt{2}}$

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

$\cos(45-30)=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos(15^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

Compare your answer with the correct one above

Question

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

$\cos(105^\circ)$ .

Answer

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 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.

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

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

$\cos(60+45)=\cos(60)\cos(45)-\sin(60)\sin(45)$

$\cos(60+45)=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}$

$\cos(60+45)=\frac{1-\sqrt{3}}{2\sqrt{2}}$

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

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

$\cos(60+45)=\frac{\sqrt{2}-\sqrt{6}}{4}$

$\cos(105^\circ)=\frac{\sqrt{2}-\sqrt{6}}{4}$

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