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Unit 5 · Analytic Trigonometry
5.1 Using
Fundamental Identities
Objective: Use reciprocal, quotient, Pythagorean, cofunction, and even/odd identities to simplify, rewrite, and factor trigonometric expressions.
Equivalent forms Reciprocal identities Quotient identities Pythagorean identities Rewrite before simplifying
Launch

Identity Family Sort

At your board: Sort the 9 cards into 3 families of expressions that are always equivalent.

You must be able to explain why each card belongs in its family. Do not just group by appearance.

Sort into 3 families
Name the connection
Defend one grouping
Create 1 new card
Card A
\[\tan x\]
Card B
\[\dfrac{\sin x}{\cos x}\]
Card C
\[\dfrac{1}{\cot x}\]
Card D
\[\csc x\]
Card E
\[\dfrac{1}{\sin x}\]
Card F
\[\dfrac{\sec x}{\tan x}\]
Card G
\[1\]
Card H
\[\sin^2 x + \cos^2 x\]
Card I
\[\sec^2 x - \tan^2 x\]

Debrief prompt: Which pair looked the least alike at first, but turned out to be in the same family?

Reference

Trigonometric Identities

Reciprocal Identities
\[\sin\theta = \dfrac{1}{\csc\theta}\]
\[\csc\theta = \dfrac{1}{\sin\theta}\]
\[\cos\theta = \dfrac{1}{\sec\theta}\]
\[\sec\theta = \dfrac{1}{\cos\theta}\]
\[\tan\theta = \dfrac{1}{\cot\theta}\]
\[\cot\theta = \dfrac{1}{\tan\theta}\]
These also work with powers, such as \(\sin^2\theta = \dfrac{1}{\csc^2\theta}\).
Quotient Identities
\[\tan\theta = \dfrac{\sin\theta}{\cos\theta}\]
\[\cot\theta = \dfrac{\cos\theta}{\sin\theta}\]
Pythagorean Identities
\[\sin^2\theta + \cos^2\theta = 1\]
\[1 + \tan^2\theta = \sec^2\theta\]
\[1 + \cot^2\theta = \csc^2\theta\]
Cofunction Identities
\[\sin\theta = \cos\left(\dfrac{\pi}{2} - \theta\right)\]
\[\cos\theta = \sin\left(\dfrac{\pi}{2} - \theta\right)\]
\[\tan\theta = \cot\left(\dfrac{\pi}{2} - \theta\right)\]
\[\cot\theta = \tan\left(\dfrac{\pi}{2} - \theta\right)\]
\[\sec\theta = \csc\left(\dfrac{\pi}{2} - \theta\right)\]
\[\csc\theta = \sec\left(\dfrac{\pi}{2} - \theta\right)\]
\(\theta\) is an acute angle.
Reference

Even / Odd Identities

  • Even functions: the output stays the same when the input changes sign, so \(\,f(-x) = f(x)\,\).
  • Odd functions: the output switches sign when the input changes sign, so \(\,f(-x) = -f(x)\,\).
Even Functions
  • \[\cos(-x) = \cos x\]
  • \[\sec(-x) = \sec x\]
Odd Functions
  • \[\sin(-x) = -\sin x\]
  • \[\tan(-x) = -\tan x\]
  • \[\csc(-x) = -\csc x\]
  • \[\cot(-x) = -\cot x\]
Example 1

Simplify the expression

\[\sin^2 x\sec x - \sec x\]
Strategy: factor first, then use a Pythagorean identity to reduce the bracket.
Factor
\[\sec x(\sin^2 x - 1)\]
Pythagorean Identity
\[\sec x(\sin^2 x - 1) = \sec x(-\cos^2 x)\]
Use \(\sin^2 x - 1 = -\cos^2 x\).
Answer
\[\dfrac{-\cos^2 x}{\cos x} = -\cos x\]
Example 2

Rewrite in an equivalent form

\[\sin^2 x - \cos^2 x\]
Strategy: replace one squared trig function with an equivalent expression involving the other.
Substitute
\[\sin^2 x - (1 - \sin^2 x)\]
Answer
\[2\sin^2 x - 1\]
An equivalent answer is \(1 - 2\cos^2 x\).
Example 3

Factor the expression

\[6\cot^2\theta - 13\cot\theta + 5\]
Strategy: let \(u = \cot\theta\) so the trinomial behaves like an ordinary quadratic.
Substitution
\[\text{Let } u = \cot\theta \quad \Rightarrow \quad 6u^2 - 13u + 5\]
Find Factors
\[6 \cdot 5 = 30,\quad -10 + (-3) = -13\]
Use \(-10\) and \(-3\) to split the middle term.
Rewrite Middle Term
\[6u^2 - 10u - 3u + 5\]
Group
\[2u(3u - 5) - 1(3u - 5)\]
Factor
\[(2u - 1)(3u - 5)\]
Answer
\[(2\cot\theta - 1)(3\cot\theta - 5)\]
Example 4

Factor after rewriting

\[\cot^2 x - 2\csc x - 7\]
Strategy: rewrite \(\cot^2 x\) using a Pythagorean identity so every term involves \(\csc x\).
Pythagorean Identity
\[\cot^2 x = \csc^2 x - 1 \quad \Rightarrow \quad \csc^2 x - 1 - 2\csc x - 7\]
Combine
\[\csc^2 x - 2\csc x - 8\]
Answer
\[(\csc x - 4)(\csc x + 2)\]
Example 5

Simplify the expression

\[\csc x\sec x - \sin x\sec x\]
Strategy: factor, convert to sine and cosine, then use a Pythagorean identity to create cancellation.
Factor
\[\sec x(\csc x - \sin x)\]
Reciprocal Identities
\[\dfrac{1}{\cos x}\left(\dfrac{1}{\sin x} - \sin x\right) = \dfrac{1}{\cos x} \cdot \dfrac{1 - \sin^2 x}{\sin x}\]
Pythagorean Identity
\[\dfrac{1}{\cos x} \cdot \dfrac{\cos^2 x}{\sin x}\]
Replace \(1 - \sin^2 x\) with \(\cos^2 x\).
Answer
\[\dfrac{\cos x}{\sin x} = \cot x\]
Example 6

Rewrite without a fraction

\[\text{Rewrite } \dfrac{\sin^2 x}{1 + \cos x} \text{ without a fraction.}\]
Strategy: factor \(\sin^2 x\) as a difference of squares so the denominator appears as a common factor.
Pythagorean Identity
\[\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)\]
Substitute
\[\dfrac{(1 - \cos x)(1 + \cos x)}{1 + \cos x}\]
Answer
\[1 - \cos x\]
Example 7

Trig substitution

\[\text{Use } x = 9\cos\theta,\; 0 < \theta < \dfrac{\pi}{2}, \text{ to write } \sqrt{81 - x^2} \text{ as a trig function of } \theta.\]
Strategy: substitute first, use a Pythagorean identity on the radicand, then use the domain to choose the correct sign.
Substitute
\[x = 9\cos\theta \Rightarrow x^2 = 81\cos^2\theta\]
Simplify Radicand
\[81 - x^2 = 81 - 81\cos^2\theta = 81(1 - \cos^2\theta)\]
Pythagorean Identity
\[81(1 - \cos^2\theta) = 81\sin^2\theta\]
Use \(1 - \cos^2\theta = \sin^2\theta\).
Answer
\[\sqrt{81 - x^2} = \sqrt{81\sin^2\theta} = 9\sin\theta\]
This is positive because \(0 < \theta < \dfrac{\pi}{2}\).
Summary

Strategies for Simplifying

1
Factor out a common term
Look for a shared factor across all terms before choosing an identity. Example 1 and Example 5 start that way.
2
Apply a Pythagorean identity
Use \(\sin^2 x + \cos^2 x = 1\) and its rearrangements to replace one squared trig expression with another.
3
Rewrite in sine and cosine
Convert \(\sec\), \(\csc\), \(\tan\), and \(\cot\) when that reveals cancellation or common structure.
4
Use \(u\)-substitution for factoring
If a trig expression repeats, rename it first so the algebra becomes a familiar polynomial.
5
Factor a difference of squares
Expressions like \(1 - \cos^2 x\) can be factored further when you need a cancellation target.
6
Look for the first move that reduces complexity
Most problems use more than one idea, so the best first step is the one that makes the expression simpler.
Most problems combine two or more of these moves. The hard part is spotting the first productive rewrite.