14-1 Trigonometric Identities Study Guide

Learning Objectives and Standards

The primary objective of this lesson is to verify trigonometric identities using algebraic techniques and fundamental trigonometric relationships.

Common Core State Standards
  • F-TF.C.8: Prove the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and use it to find sin⁡(θ)\sin(\theta), cos⁡(θ)\cos(\theta), or tan⁡(θ)\tan(\theta), given sin⁡(θ)\sin(\theta), cos⁡(θ)\cos(\theta), or tan⁡(θ)\tan(\theta) and the quadrant of the angle.
  • Mathematical Practices (MP): The lesson incorporates MP 1, MP 2, MP 3, and MP 4.
Core Goals
  • Use Algebra to simplify Trigonometric Expressions.
  • Establish familiarity with basic variations of fundamental identities.

Definitions of Equations and Identities

Identically Equal

Two functions ff and gg are considered identically equal if for every value of xx for which both functions are defined, the equation f(x)=g(x)f(x) = g(x) holds true.

Identity

An equation that is identically equal is referred to as an identity. These are true for all defined values in the domain.

Conditional Equation

An equation that is not an identity is called a conditional equation. These equations are only true for specific values of the variable.

Examples of Identities and Conditional Equations

Identities
  • Algebraic: (x+1)2=x2+2x+1(x + 1)^2 = x^2 + 2x + 1
  • Pythagorean: sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1
  • Reciprocal: csc⁡(x)=1sin⁡(x)\csc(x) = \frac{1}{\sin(x)}
Conditional Equations
  • Linear: 2x+5=02x + 5 = 0 is true only if x=−52x = -\frac{5}{2}.
  • Trigonometric Roots: sin⁡(x)=0\sin(x) = 0 is true only if x=kπx = k\pi, where kk is an integer.
  • Trigonometric Substitution: sin⁡(x)=cos⁡(x)\sin(x) = \cos(x) is true only if x=π4+2kπx = \frac{\pi}{4} + 2k\pi or x=5π4+2kπx = \frac{5\pi}{4} + 2k\pi, where kk is an integer.

Fundamental Trigonometric Identities

These identities are considered the "basic" set and should be known fluently, including their minor variations (e.g., recognizing sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) as a variation of the Pythagorean identity).

Quotient Identities
  • tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
  • cot⁡(θ)=cos⁡(θ)sin⁡(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}
Reciprocal Identities
  • csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}
  • sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}
  • cot⁡(θ)=1tan⁡(θ)\cot(\theta) = \frac{1}{\tan(\theta)}
Pythagorean Identities
  • sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1
  • tan⁡2(θ)+1=sec⁡2(θ)\tan^2(\theta) + 1 = \sec^2(\theta)
  • cot⁡2(θ)+1=csc⁡2(θ)\cot^2(\theta) + 1 = \csc^2(\theta)
Even-Odd Identities
  • sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta)
  • csc⁡(−θ)=−csc⁡(θ)\csc(-\theta) = -\csc(\theta)
  • cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta)
  • sec⁡(−θ)=sec⁡(θ)\sec(-\theta) = \sec(\theta)
  • tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta)
  • cot⁡(−θ)=−cot⁡(θ)\cot(-\theta) = -\cot(\theta)

Algebraic Techniques to Simplify Trigonometric Expressions

Example 1: Rewriting in terms of Sine and Cosine

Simplify cot⁡(θ)csc⁡(θ)\frac{\cot(\theta)}{\csc(\theta)}:

  1. Express terms using sine and cosine: cos⁡(θ)sin⁡(θ)1sin⁡(θ)\frac{\frac{\cos(\theta)}{\sin(\theta)}}{\frac{1}{\sin(\theta)}}
  2. Multiply by the reciprocal: cos⁡(θ)sin⁡(θ)×sin⁡(θ)1\frac{\cos(\theta)}{\sin(\theta)} \times \frac{\sin(\theta)}{1}
  3. Result: cos⁡(θ)\cos(\theta)
Example 2: Multiplying by a Conjugate

Show that cos⁡(θ)1+sin⁡(θ)=1−sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{1 + \sin(\theta)} = \frac{1 - \sin(\theta)}{\cos(\theta)} by multiplying the numerator and denominator by (1−sin⁡(θ))\left(1 - \sin(\theta)\right).

  1. Step: cos⁡(θ)(1+sin⁡(θ))×(1−sin⁡(θ))(1−sin⁡(θ))\frac{\cos(\theta)}{(1 + \sin(\theta))} \times \frac{(1 - \sin(\theta))}{(1 - \sin(\theta))}
  2. Expand the denominator: cos⁡(θ)(1−sin⁡(θ))1−sin⁡2(θ)\frac{\cos(\theta)(1 - \sin(\theta))}{1 - \sin^2(\theta)}
  3. Use Pythagorean identity (1−sin⁡2(θ)=cos⁡2(θ)1 - \sin^2(\theta) = \cos^2(\theta)): cos⁡(θ)(1−sin⁡(θ))cos⁡2(θ)\frac{\cos(\theta)(1 - \sin(\theta))}{\cos^2(\theta)}
  4. Simplify: 1−sin⁡(θ)cos⁡(θ)\frac{1 - \sin(\theta)}{\cos(\theta)}
Example 3: Rewriting as a Single Ratio

Simplify 1+sin⁡(u)sin⁡(u)+cot⁡(u)−cos⁡(u)cos⁡(u)\frac{1 + \sin(u)}{\sin(u)} + \frac{\cot(u) - \cos(u)}{\cos(u)}:

  1. Expand fractions: 1sin⁡(u)+sin⁡(u)sin⁡(u)+cot⁡(u)cos⁡(u)−cos⁡(u)cos⁡(u)\frac{1}{\sin(u)} + \frac{\sin(u)}{\sin(u)} + \frac{\cot(u)}{\cos(u)} - \frac{\cos(u)}{\cos(u)}
  2. Substitute identities: csc⁡(u)+1+cos⁡(u)sin⁡(u)cos⁡(u)−1\csc(u) + 1 + \frac{\frac{\cos(u)}{\sin(u)}}{\cos(u)} - 1
  3. Simplify constants and ratios: csc⁡(u)+1sin⁡(u)\csc(u) + \frac{1}{\sin(u)}
  4. Combined: csc⁡(u)+csc⁡(u)=2csc⁡(u)\csc(u) + \csc(u) = 2\csc(u)Note: An alternative method involving a common denominator of sin⁡(u)cos⁡(u)\sin(u)\cos(u) yields 2cos⁡(u)sin⁡(u)cos⁡(u)=2sin⁡(u)=2csc⁡(u)\frac{2\cos(u)}{\sin(u)\cos(u)} = \frac{2}{\sin(u)} = 2\csc(u).
Example 4: Factoring

Simplify sin⁡2(v)−1tan⁡(v)sin⁡(v)−tan⁡(v)\frac{\sin^2(v) - 1}{\tan(v)\sin(v) - \tan(v)}:

  1. Factor the numerator (Difference of Squares): (sin⁡(v)+1)(sin⁡(v)−1)(\sin(v) + 1)(\sin(v) - 1)
  2. Factor the denominator (Greatest Common Factor): tan⁡(v)(sin⁡(v)−1)\tan(v)(\sin(v) - 1)
  3. Cancel common factor (sin⁡(v)−1)(\sin(v) - 1): sin⁡(v)+1tan⁡(v)\frac{\sin(v) + 1}{\tan(v)}

Questions & Discussion

Warm-Up 1: True or False: sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta).

  • Answer: True. This is a basic algebraic rearrangement of the Pythagorean Identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.

Warm-Up 2: True or False: sin⁡(−θ)+cos⁡(−θ)=cos⁡(θ)−sin⁡(θ)\sin(-\theta) + \cos(-\theta) = \cos(\theta) - \sin(\theta).

  • Answer: True. By using Even-Odd Identities, sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta) and cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta). Therefore, −sin⁡(θ)+cos⁡(θ)-\sin(\theta) + \cos(\theta) is equivalent to cos⁡(θ)−sin⁡(θ)\cos(\theta) - \sin(\theta).

Practice Problem 1: Rewrite cot⁡(θ)⋅sec⁡(θ)\cot(\theta) \cdot \sec(\theta) in terms of sine and cosine functions.

  • Work: cos⁡(θ)sin⁡(θ)×1cos⁡(θ)\frac{\cos(\theta)}{\sin(\theta)} \times \frac{1}{\cos(\theta)}
  • Result: 1sin⁡(θ)=csc⁡(θ)\frac{1}{\sin(\theta)} = \csc(\theta)

Practice Problem 2: Multiply sin⁡(θ)1+cos⁡(θ)\frac{\sin(\theta)}{1 + \cos(\theta)} by 1−cos⁡(θ)1−cos⁡(θ)\frac{1 - \cos(\theta)}{1 - \cos(\theta)}.

  • Work: sin⁡(θ)(1−cos⁡(θ))1−cos⁡2(θ)=sin⁡(θ)(1−cos⁡(θ))sin⁡2(θ)\frac{\sin(\theta)(1 - \cos(\theta))}{1 - \cos^2(\theta)} = \frac{\sin(\theta)(1 - \cos(\theta))}{\sin^2(\theta)}
  • Result: 1−cos⁡(θ)sin⁡(θ)\frac{1 - \cos(\theta)}{\sin(\theta)}

Practice Problem 3: Rewrite over a common denominator: 11−cos⁡(v)+11+cos⁡(v)\frac{1}{1 - \cos(v)} + \frac{1}{1 + \cos(v)}.

  • Work: (1+cos⁡(v))+(1−cos⁡(v))(1−cos⁡(v))(1+cos⁡(v))=21−cos⁡2(v)\frac{(1 + \cos(v)) + (1 - \cos(v))}{(1 - \cos(v))(1 + \cos(v))} = \frac{2}{1 - \cos^2(v)}
  • Result: 2sin⁡2(v)=2csc⁡2(v)\frac{2}{\sin^2(v)} = 2\csc^2(v)

Homework Assignment

  • Reference: 14-1 Homework #1
  • Source: Textbook Page 470
  • Task: Problems 9 through 18, all.