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All Common Core: High School - Functions Resources

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Flashcards: Common Core: High School - Functions

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

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

find

$\tan(\theta)$ ?

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

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

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

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

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

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Flashcards: Common Core: High School - Functions

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

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

find

$\tan(\theta)$ ?

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

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 $\theta$ .

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

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

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

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.

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

Flashcards: Common Core: High School - Functions

$g(x)=\frac{x}{x^2-x}$

What is the domain of the function ?

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq -1 \end{Bmatrix}$

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\leq 1 \end{Bmatrix}$

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\geq 1 \end{Bmatrix}$

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq 1 \end{Bmatrix}$

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq 0 \end{Bmatrix}$

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Flashcards: Common Core: High School - Functions

$g(x)=\frac{x}{x^2-x}$

What is the domain of the function ?

Your answer:

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq -1 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\leq 1 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\geq 1 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq 0 \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq 1 \end{Bmatrix}$

Explanation:

This question is testing the concept and understanding of a function's domain. It is important to recall that domain can be identified graphically or algebraically. Graphically, domain contains the x-values of the function where as range, contains the y-values that span the image of the function. Algebraically, domain is known as the input values x of a function that then results in the y outputs. In other words, the input values when placed into the function results in the y values creating an (x, y) pair. It is also critical to understand that there are two important restrictions on domain. These restrictions occur when the x variable is in the denominator of a function or under a radical. This is because a fraction does not exist when there is a zero in the denominator; if under the radical is negative it results in imaginary values.

For the purpose of Common Core Standards, domain and range fall within the Cluster A of the function and use of function notation concept (CCSS.Math.content.HSF-IF.A).

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 the function and what the question is asking.

$g(x)=\frac{x}{x^2-x}$

Find the domain (x values) of the function. This includes identifying points where x values do not result in a y value. At these points the domain does not exist.

Step 2. Discuss the options to solve the problem.

I. Graphically plot the function by computer/technology resource. Then interpret the graph.

II. Create a table of (x, y) pairs and plot the points to create the graph. Then interpret the graph.

III. Algebraically find the x values that do not exist, these will be areas that are not in the domain.

For this particular function let's use the third option to simplify the function and then use the first option to graph and solve for the domain.

$g(x)=\frac{x}{x^2-x}$

To simplify this function factor out a common factor of in the numerator and denominator.

$g(x)=\frac{x(1)}{x(x-1)}=\frac{1}{x-1}$

Any time there is a fraction set the denominator equal to zero and solve for x, this will be a value that x cannot equal since it is not in the domain.

Fractions:

$y=\frac{1}{a-x} \rightarrow a-x\neq 0\rightarrow a\neq x$

Step 3: Use algebraic technique along with graphing the function to solve the problem.

$\\ x-1\neq 0 \\x+1-1\neq0+1 \\x\neq1$

Graph to verify solution.

Screen shot 2016 01 07 at 2.02.35 pm

The graph above shows the function to have a vertical asymptote at x=1 therefore it is not in the domain.

Thus the solution to this problem is,

$\begin{Bmatrix} x\ \epsilon \ \mathbb{R}|\ x\neq 1 \end{Bmatrix}$

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All Common Core: High School - Functions Resources

Flashcards: Common Core: High School - Functions

Flashcards: Common Core: High School - Functions

Flashcards: Common Core: High School - Functions

Flashcards: Common Core: High School - Functions

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