Midterm 2 Outline and Reminders:

MEMORIAL UNIVERSITY OF NEWFOUNDLAND EXAM DETAILS

  • Course: Mathematics 1090(081)

  • Department: Mathematics and Statistics

  • Exam Type: Midterm II

  • Year: Fall 2018

  • Total Marks: 100 marks

  • Duration: 120 minutes

    • Time Management Suggestion: Allocate approximately 1 mark per minute for each question, leaving around 20 minutes at the end for reviewing your answers and checking for errors.

  • Instructions: These rules are crucial for exam success.

    • Always show all your work clearly to receive the best possible marks. Even if your final answer is incorrect, partial marks can be awarded for correct steps.

    • The exam is closed-book: no notes, textbooks, calculators, or any electronic devices are permitted. Ensure you are familiar with manual calculations and fundamental concepts.

    • All work, including any rough calculations, must be done directly on the exam paper. No scrap paper is allowed, so organize your workspace on the provided paper.

GENERAL EXAM STRUCTURE

  • Question Breakdown: The exam consists of a variety of questions covering different topics.

    • Total Questions: 10 individual questions.

    • Marks allocated per question varies significantly, so pay attention to the weight of each question when managing your time.

QUESTIONS: Detailed Breakdown with Strategies
Question #1 (6 marks): Polynomial Factor and Remainder Theorem

  • Task: Find the value of k in a given polynomial expression, P(x).

  • Problem-Solving Tip: This question tests your understanding of the Factor Theorem and the Remainder Theorem, which are fundamental for working with polynomials.

    a) Determine k if P(x) has a factor of (x minus 1).

    - **Strategy**: Recall the Factor Theorem: if (x minus r) is a factor of P(x), then P(r) must equal zero. In this case, since (x minus 1) is a factor, you set P(1) = 0 and solve the resulting equation for k.

    b) Determine k if P(x) has a remainder of 12 when divided by (x minus 2).

    - **Strategy**: Recall the Remainder Theorem: if P(x) is divided by (x minus r), the remainder is P(r). Here, since the remainder is 12 when divided by (x minus 2), you set P(2) = 12 and solve for k.
Question #2 (5 marks): Constructing a Polynomial from Roots

  • Task: Identify a polynomial P(x) of the lowest possible degree that satisfies specific root conditions.

  • Problem-Solving Tip: Remember that a root with a certain multiplicity means the corresponding factor is raised to that power. The given point helps determine the leading coefficient.

    • It has a root of minus 2 with multiplicity 2.

      • Strategy: If r is a root with multiplicity m, then (x minus r) raised to the power of m is a factor of the polynomial. For a root of minus 2 with multiplicity 2, the factor will be (x minus (minus 2))^2, which simplifies to (x plus 2)^2.

    • P(0) = minus 8.

      • Strategy: After forming the basic polynomial structure P(x) = C * (x plus 2)^2 (where C is the leading coefficient), use this condition to find C. Substitute x = 0 into your polynomial expression and set it equal to minus 8, then solve for C. For example, P(0) = C * (0 plus 2)^2 = C * (2)^2 = 4C. So, 4C = minus 8 means C = minus 2.

Question #3 (10 marks): Factoring Polynomials and Finding Roots

  • Task: Given the polynomial 2x^4 + 3x^3 - 11x^2 - 19x + 6, perform a series of actions related to its roots and factors.

  • Problem-Solving Tip: The Rational Root Theorem is your best friend here. Synthetic division is the practical tool to test potential roots and factor the polynomial step-by-step.

    a) List all potential rational roots/zeros of the polynomial.

    - **Strategy**: Use the Rational Root Theorem. Potential rational roots are in the form `p/q`, where `p` is a factor of the constant term (6) and `q` is a factor of the leading coefficient (2). List all positive and negative factors of 6 (which are plus/minus 1, plus/minus 2, plus/minus 3, plus/minus 6). List all positive and negative factors of 2 (which are plus/minus 1, plus/minus 2). Then, form all possible fractions `p/q` to get your list of potential rational roots.

    b) Fully factor the polynomial.

    - **Strategy**: Use synthetic division. Test the potential rational roots found in part (a). If a root `r` yields a remainder of zero, then `(x minus r)` is a factor. Repeat the process with the resulting depressed polynomial until you reach a quadratic expression, which can then be factored using standard methods (factoring, quadratic formula, etc.).

    c) List all actual rational roots/zeros of the polynomial.

    - **Strategy**: Once you have fully factored the polynomial, the actual rational roots are the values of x that make each factor equal to zero.
Question #4 (6 marks): Truth Value of Statements

  • Task: Determine whether given mathematical statements are true or false, providing a brief explanation for your answer.

  • Problem-Solving Tip: This section assesses your foundational understanding of mathematical definitions, theorems, and properties, likely related to polynomials, functions, or basic algebra. Read each statement carefully and base your explanation on established mathematical principles.

Question #5 (9 marks): Trigonometric Definitions

  • Task: This question focuses on fundamental trigonometric definitions, often involving diagrams.

  • Problem-Solving Tip: Visual aids like diagrams are excellent for remembering these. Be precise with your definitions.

    a) Complete blanks in a diagram.

    - **Strategy**: Be prepared to label angles, sides (opposite, adjacent, hypotenuse), or coordinates (x, y, r) in a right-angle triangle or unit circle context.

    b) Label the diagram.

    - **Strategy**: Similar to (a), ensure all relevant parts of the provided diagram are correctly identified and labeled.

    c) Define trigonometric functions using:

    - **Right-angle triangle definitions**
    - **Strategy**: Remember the acronym SOH CAH TOA:

    - **Sine (theta)** = Opposite / Hypotenuse

    - **Cosine (theta)** = Adjacent / Hypotenuse

    - **Tangent (theta)** = Opposite / Adjacent

Also define their reciprocals:

    - **Cosecant (theta)** = Hypotenuse / Opposite (reciprocal of sine)

    - **Secant (theta)** = Hypotenuse / Adjacent (reciprocal of cosine)

    - **Cotangent (theta)** = Adjacent / Opposite (reciprocal of tangent)

    - **Definitions for sine and cosine.**

    - **Strategy**: These typically refer to the unit circle or general angle definitions where a point (x, y) is on the terminal arm of an angle theta, and r is the distance from the origin to (x, y).

    - **Sine (theta)** = y / r

    - **Cosine (theta)** = x / r
Question #6 (9 marks): Evaluating Trigonometric Expressions

  • Task: Evaluate various trigonometric expressions without any restrictions on the method used.

  • Problem-Solving Tip: Mastery of common angle values (0, 30, 45, 60, 90 degrees or 0, pi/6, pi/4, pi/3, pi/2 radians) and their trigonometric function values is key. Understand which quadrant the angle lies in to determine the sign of the function.

    a) 4 multiplied by the sine of 3 theta

    b) 13 multiplied by the cotangent of 4 theta

    c) 7 multiplied by the secant of 6 theta

    • Strategy: For each part, first evaluate the trigonometric function (e.g., sine of 3 theta) for the given angle, then multiply by the constant. If specific numerical angles are provided for theta, calculate the inner angle first.

Question #7 (16 marks): Trigonometric Simplification and Identities

  • Task: This section combines simplification of expressions with proving trigonometric identities and solving trigonometric equations.

  • Problem-Solving Tip: Express everything in terms of sine and cosine for simplification. For proofs, work on one side until it matches the other. For solving, isolate the trigonometric function.

    a) Simplify: cosecant theta multiplied by tangent theta.

    - **Strategy**: Convert cosecant and tangent into their sine and cosine equivalents: `cosecant theta = 1 / sine theta` and `tangent theta = sine theta / cosine theta`. Then multiply and simplify the expression.

    b) Prove the identity: tangent squared theta plus secant squared theta equals 3 secant squared theta.

    - **Strategy**: To prove an identity, typically you start with one side (usually the more complex one) and use known identities to transform it step-by-step until it looks identical to the other side. A common Pythagorean identity is `tangent squared theta plus 1 equals secant squared theta`. If the identity given in the problem is `tan^2(theta) + sec^2(theta) = 3sec^2(theta)`, this can be rewritten as `tan^2(theta) = 2sec^2(theta)`. You would typically work to show if the left side equals the right side using other known identities.

    c) Solve for theta in 3 times sine squared theta minus 2 equals 0 if given specific conditions.

    - **Strategy**: First, isolate `sine squared theta`. You'll get `sine squared theta = 2/3`. Then take the square root of both sides to get `sine theta = plus/minus sqrt(2/3)`. Identify the angles (theta) within the specified domain or interval that satisfy this sine value. Remember that there will usually be two angle solutions per cycle for both positive and negative sine values.
Question #8 (12 marks): Identity Utilization

  • Task: Apply trigonometric identities to find expressions or simplify terms.

  • Problem-Solving Tip: Practice recognizing patterns that align with angle sum/difference, double angle, or other key identities. Sometimes, you might need to manipulate the given expression algebraically before applying an identity.

    a) Find an expression involving only sine and cosine.

    - **Strategy**: This usually means converting any secant, cosecant, tangent, or cotangent functions into their base sine and cosine forms and simplifying.

    b) Simplify a compound trigonometric expression.

    - **Strategy**: Look for opportunities to use sum and difference identities (e.g., sine of (A plus B), cosine of (A minus B)), or double angle identities (e.g., sine of (2A), cosine of (2A)). Algebraic factoring or combining terms might also be necessary.
Question #9 (18 marks): Advanced Trigonometric Evaluation

  • Task: Given some trigonometric context or information, perform calculations involving different trigonometric functions and angles.

  • Problem-Solving Tip: Often, you'll need to find one trigonometric value first (e.g., sine or cosine) and then use Pythagorean identities (like sine squared theta plus cosine squared theta equals 1) to find others, or use double angle formulas.

    a) Find tangent theta.

    - **Strategy**: If you know sine theta and cosine theta, `tangent theta = sine theta / cosine theta`. If you're given a point (x, y) on the terminal arm of theta, `tangent theta = y / x`.

    b) Evaluate tangent of k theta.

    - **Strategy**: This often involves applying double angle or multiple angle formulas for tangent, depending on the value of `k`. For instance, if `k = 2`, use the tangent double angle formula.

    c) Calculate cosine of 2 theta.

    - **Strategy**: There are three common double angle formulas for cosine:
    - `cosine (2 theta) = cosine squared theta minus sine squared theta`

- `cosine (2 theta) = 2 times cosine squared theta minus 1`

- `cosine (2 theta) = 1 minus 2 times sine squared theta`

    Choose the most convenient formula based on whether you already know the value of sine theta or cosine theta.

Question #10 (6 marks): Expression Manipulation and Proof

  • Task: This question involves algebraic manipulation of trigonometric expressions and proving relations.

  • Problem-Solving Tip: For manipulation, treat trigonometric functions as variables if it helps, but remember their underlying identities. For proofs, show clear, step-by-step transformations.

    a) Given the expression, manipulate it according to the given instructions.

    - **Strategy**: This is a general task. It might involve factoring, expanding, combining fractions, or converting to a common base (like sine and cosine) to simplify the expression.

    b) Prove a relation involving cosine and sine functions and their respective angles.

    - **Strategy**: Similar to proving identities in Question 7, pick one side of the relation and use algebraic steps and trigonometric identities to transform it until it is identical to the other side. Think about using angle sum/difference formulas if there are different angles involved.
ADDITIONAL MATERIAL: Key Identities and Applications

  • Trigonometric Identities: These are essential formulas you must know by heart; they are your tools for solving trigonometric problems.

    • Important formulas and identities for solving problems:

      • Basic Pythagorean Identity: sine squared theta plus cosine squared theta equals 1. From this, you can also derive tangent squared theta plus 1 equals secant squared theta and 1 plus cotangent squared theta equals cosecant squared theta.

      • Reciprocal Identities: cosecant theta = 1 / sine theta, secant theta = 1 / cosine theta, cotangent theta = 1 / tangent theta.

      • Quotient Identities: tangent theta = sine theta / cosine theta, cotangent theta = cosine theta / sine theta.

      • Angle Sum and Difference Formulas: These allow you to find the sine, cosine, or tangent of a sum or difference of two angles (e.g., sine of (A plus B), cosine of (A minus B)).

      • Double Angle Formulas: These are used to find the sine, cosine, or tangent of twice an angle (e.g., sine of (2 theta), cosine of (2 theta)).

  • Kite Problem: A practical application of trigonometry, often encountered in word problems.

    • Mathematical application involving altitude based on rope length and angle of elevation:

      • Given: A kite attached to a 100-meter rope at an angle of elevation of pi over 6 radians (which is 30 degrees). The task is to solve for the height (altitude) of the kite.

      • Strategy: Draw a right-angle triangle. The kite string forms the hypotenuse (100 meters). The height of the kite is the side opposite the angle of elevation. The angle of elevation is given as pi over 6 radians (30 degrees). The trigonometric function that relates opposite and hypotenuse is sine. So, sine (angle of elevation) = height / rope length. Therefore, height = rope length * sine (angle of elevation). Substitute the given values: height = 100 * sine (pi over 6). Since sine (pi over 6) (or sine (30 degrees)) equals 1/2, the height would be 100 * (1/2) = 50 meters.