Final Review: Advanced Algebra and Trigonometry

As we delve into the intricacies of Advanced Algebra and Introductory Trigonometry, let's refresh our minds and sharpen our skills through a comprehensive review of key concepts, formulas, and techniques. This final review will cover essential topics, complete with sample problems and solutions, to help you solidify your understanding and excel in your mathematical journey.

1. Functions and Graphs

Linear Functions

A linear function can be represented in the slope-intercept form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

Example Problem: Find the slope and y-intercept of the line given by the equation \( 3y - 6x = 9 \).

Solution:

  1. Rearrange the equation to slope-intercept form:
    • \( 3y = 6x + 9 \)
    • \( y = 2x + 3 \)

Here, the slope \( m = 2 \) and the y-intercept \( b = 3 \).

Quadratic Functions

A quadratic function is given by the standard form \( f(x) = ax^2 + bx + c \). The vertex form is useful for graphing and is expressed as \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex.

Example Problem: Convert \( f(x) = 2x^2 + 8x + 6 \) to vertex form.

Solution:

  1. Factor out \( 2 \) from the quadratic term:
    • \( f(x) = 2(x^2 + 4x) + 6 \)
  2. Complete the square:
    • \( f(x) = 2((x + 2)^2 - 4) + 6 \)
    • \( f(x) = 2(x + 2)^2 - 8 + 6 \)
    • \( f(x) = 2(x + 2)^2 - 2 \)

The vertex is \( (-2, -2) \).

2. Exponents and Polynomials

Laws of Exponents

Understanding the laws of exponents is crucial, especially in simplifying expressions and solving equations.

Example Problem: Simplify \( (x^3y^2)^2 \cdot x^{-2} \).

Solution:

  1. Apply the power of a power rule:
    • \( (x^3y^2)^2 = x^6y^4 \)
  2. Now multiply by \( x^{-2} \):
    • \( x^6y^4 \cdot x^{-2} = x^{6-2}y^4 = x^4y^4 \)

Polynomial Division

Polynomial long division is similar to numerical long division.

Example Problem: Divide \( x^3 + 2x^2 - x + 3 \) by \( x + 1 \).

Solution:

  1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
  2. Multiply \( x^2 \) by \( (x + 1) \) and subtract:
    • \( (x^3 + 2x^2 - x + 3) - (x^3 + x^2) = x^2 - x + 3 \)
  3. Repeat with \( x^2 - x + 3 \):
    • Divide \( x^2 \) by \( x \) to get \( x \).
    • Subtract \( (x^2 + x) \) from \( (x^2 - x + 3) \):
      • Result is \( -2x + 3 \).
  4. Finally, divide \( -2x \) by \( x \) to get \(-2\) and subtract:
    • \( (-2x + 3) - (-2x - 2) = 5 \).

The complete result is \( x^2 + x - 2 + \frac{5}{x+1} \).

3. Trigonometric Functions

Basic Trigonometric Ratios

The primary trigonometric ratios are sine, cosine, and tangent, defined as follows for a right triangle with angle \( \theta \):

  • \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
  • \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

Example Problem: If \( \theta \) is an angle in a right triangle with an opposite side of 4 and a hypotenuse of 5, find \( \sin(\theta) \) and \( \cos(\theta) \).

Solution:

  1. For sine:
    • \( \sin(\theta) = \frac{4}{5} \)
  2. Find the adjacent side using the Pythagorean theorem:
    • \( a^2 + 4^2 = 5^2 \) leads to \( a^2 + 16 = 25 \) \( \Rightarrow a^2 = 9 \Rightarrow a = 3 \).
  3. For cosine:
    • \( \cos(\theta) = \frac{3}{5} \).

The Unit Circle

The unit circle is a circle of radius 1 centered at the origin \( (0, 0) \). The angle corresponds to coordinates on the circle: \( (cos(\theta), sin(\theta)) \).

Example Problem: Find the values of \( \sin \) and \( \cos \) for \( \theta = 60^\circ \).

Solution: Using the unit circle:

  • \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
  • \( \cos(60^\circ) = \frac{1}{2} \)

4. Trigonometric Identities

Pythagorean Identity

The Pythagorean identity states that \( \sin^2(x) + \cos^2(x) = 1 \).

Example Problem: Use the Pythagorean identity to find \( \sin^2(30^\circ) \).

Solution: We know \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), thus:

  • \( \sin^2(30^\circ) + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \)
  • \( \sin^2(30^\circ) + \frac{3}{4} = 1 \)
  • \( \sin^2(30^\circ) = 1 - \frac{3}{4} = \frac{1}{4} \)
  • Therefore, \( \sin(30^\circ) = \frac{1}{2} \).

Sum and Difference Formulas

The sum and difference formulas for sine and cosine are essential for solving many trigonometric problems.

Example Problem: Calculate \( \sin(45^\circ + 30^\circ) \).

Solution: Using the formula:

  • \( \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) \).
  • Thus, \( \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) \)
  • Knowing \( \sin(45^\circ) = \frac{\sqrt{2}}{2}, \cos(30^\circ) = \frac{\sqrt{3}}{2}, \cos(45^\circ) = \frac{\sqrt{2}}{2}, \sin(30^\circ) = \frac{1}{2} \):

\[ \sin(45^\circ + 30^\circ) = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}. \]

Conclusion

Through this final review, we've revisited core topics in Advanced Algebra and Introductory Trigonometry, applying them through practical problems and solutions. Remember, practice is key to mastering these concepts. Take your time to work through more examples, and soon these principles will inherently guide you in tackling complex mathematical challenges. Best of luck in your studies!