9.3 Trigonometric Identities and Equations

9.3 Trigonometric Identities and Equations

Quick Answer: The fundamental Pythagorean identity is $\sin^2\theta + \cos^2\theta = 1$. Dividing by $\cos^2\theta$ yields $1 + \tan^2\theta = \sec^2\theta$, and dividing by $\sin^2\theta$ yields $1 + \cot^2\theta = \csc^2\theta$. When given a trigonometric ratio and quadrant, find other ratios by substituting into $\sin^2\theta + \cos^2\theta = 1$ and choosing the sign matching the quadrant. To solve trigonometric equations on $[0, 2\pi)$, factor the expression (GCF or quadratic trinomial), set each factor equal to zero, and find all corresponding angles using reference angles.

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1. The Fundamental Pythagorean Identity & Secondary Derivations (F-TF.8)

An algebraic identity is an equation that remains true for all values of the variable within the domains of the expressions involved. In trigonometry, identities allow complex expressions to be simplified and provide the mechanism to solve higher-degree trigonometric equations.

Deriving the Primary Identity: $\sin^2\theta + \cos^2\theta = 1$

Every point $P(x, y)$ on the unit circle satisfies the Cartesian circle equation:

$$x^2 + y^2 = 1$$

By the unit circle definitions established in Section 9.1, $x = \cos\theta$ and $y = \sin\theta$. Substituting these definitions into the circle equation yields:

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

This is the primary Pythagorean identity. It can be algebraically rearranged into two indispensable substitution formulas:

$$\sin^2\theta = 1 - \cos^2\theta \quad \text{and} \quad \cos^2\theta = 1 - \sin^2\theta$$

Deriving Secondary Pythagorean Identities

The two secondary Pythagorean identities are derived directly by dividing the primary identity by $\cos^2\theta$ and $\sin^2\theta$, respectively:

1. Dividing by $\cos^2\theta$ (assuming $\cos\theta \neq 0$):

   $$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$

   Applying the quotient identity $\frac{\sin\theta}{\cos\theta} = \tan\theta$ and reciprocal identity $\frac{1}{\cos\theta} = \sec\theta$:

   $$\tan^2\theta + 1 = \sec^2\theta \iff 1 + \tan^2\theta = \sec^2\theta$$

   Rearrangements: $\tan^2\theta = \sec^2\theta - 1$ and $\sec^2\theta - \tan^2\theta = 1$.

2. Dividing by $\sin^2\theta$ (assuming $\sin\theta \neq 0$):

   $$\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}$$

   Applying the quotient identity $\frac{\cos\theta}{\sin\theta} = \cot\theta$ and reciprocal identity $\frac{1}{\sin\theta} = \csc\theta$:

   $$1 + \cot^2\theta = \csc^2\theta$$

   Rearrangements: $\cot^2\theta = \csc^2\theta - 1$ and $\csc^2\theta - \cot^2\theta = 1$.

Master Summary of Trigonometric Identities

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2. Determining Exact Ratios from One Known Ratio and Quadrant

When given the value of one trigonometric function and the quadrant of angle $\theta$, all remaining five trigonometric functions can be determined algebraically using Pythagorean identities.

Systematic Solution Protocol

1. Identify the Given Value & Quadrant: Note the algebraic signs of $x$ and $y$ dictated by ASTC in that quadrant.
2. Apply the Pythagorean Identity:
   • If $\sin\theta$ is given, solve $\cos\theta = \pm\sqrt{1 - \sin^2\theta}$. Choose the $+$ or $-$ sign matching the quadrant's cosine sign.
   • If $\cos\theta$ is given, solve $\sin\theta = \pm\sqrt{1 - \cos^2\theta}$. Choose the sign matching the quadrant's sine sign.
   • If $\tan\theta$ is given, solve $\sec\theta = \pm\sqrt{1 + \tan^2\theta}$, take the reciprocal to find $\cos\theta$, and then calculate $\sin\theta = \tan\theta \cdot \cos\theta$.
3. Compute Reciprocals: Invert fractions to find $\sec\theta, \csc\theta,$ and $\cot\theta$.

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3. Algebraic Verification of Trigonometric Identities

Verifying an identity means proving that two expressions are mathematically equivalent for all defined domain values. On the Regents Examination, identity proofs must follow strict conventions:

Essential Algebraic Strategies for Identity Proofs

1. Convert to Sines and Cosines: Replace $\tan\theta, \cot\theta, \sec\theta,$ and $\csc\theta$ using their quotient and reciprocal formulas.
2. Create Common Denominators: When adding or subtracting fractions, determine the least common denominator (LCD) and combine into a single rational fraction.
3. Factor Algebraic Expressions: Look for greatest common factors (GCFs), differences of two squares ($1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$), or quadratic trinomials.
4. Multiply by the Conjugate: If a denominator contains a binomial like $1 - \sin\theta$ or $1 + \cos\theta$, multiply numerator and denominator by its conjugate ($1 + \sin\theta$ or $1 - \cos\theta$) to generate a Pythagorean identity in the denominator ($1 - \sin^2\theta = \cos^2\theta$).

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4. Solving Trigonometric Equations over $[0, 2\pi)$

Solving a trigonometric equation involves finding all angles $\theta$ or $x$ within a specified interval (typically $[0, 2\pi)$ or $[0^\circ, 360^\circ)$) that satisfy the equation.

Linear Trigonometric Equations

Isolate the trigonometric function using inverse operations, then determine the reference angle and locate all valid angles in the interval:

$$2\sin(x) - 1 = 0 \implies 2\sin(x) = 1 \implies \sin(x) = \frac{1}{2}$$

• The reference angle is $x_R = \frac{\pi}{6}$.
• Sine is positive in Quadrants I and II:
  • Quadrant I: $x = \frac{\pi}{6}$
  • Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
  • Solution Set: $\left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\}$

Quadratic Trigonometric Equations (Factoring)

When equations contain squared trigonometric terms, rearrange into standard quadratic form equal to zero ($a u^2 + b u + c = 0$, where $u = \sin x$ or $u = \cos x$):

   Original: 2cos²(x) - cos(x) - 1 = 0   ---> Substitute u = cos(x): 2u² - u - 1 = 0
   Factored: (2u + 1)(u - 1) = 0         ---> Restore: (2cos(x) + 1)(cos(x) - 1) = 0
   Branches: cos(x) = -1/2   OR   cos(x) = 1

Using Pythagorean Identities to Create a Single Function

If an equation contains mixed functions (e.g., both $\sin^2(x)$ and $\cos(x)$), use Pythagorean identities to rewrite the equation in terms of a single trigonometric function before factoring:

$$2\sin^2(x) - \cos(x) - 1 = 0$$

Substitute $\sin^2(x) = 1 - \cos^2(x)$:

$$2(1 - \cos^2(x)) - \cos(x) - 1 = 0 \implies 2 - 2\cos^2(x) - \cos(x) - 1 = 0$$ $$-2\cos^2(x) - \cos(x) + 1 = 0 \implies 2\cos^2(x) + \cos(x) - 1 = 0$$

Factor the quadratic trinomial:

$$(2\cos(x) - 1)(\cos(x) + 1) = 0$$

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5. Worked Examples

Worked Problem 1: Exact Trigonometric Values from Quadrant Conditions

Problem: Given that $\cos\theta = -\frac{4}{5}$ and $\tan\theta > 0$, algebraically determine the exact value of $\sin\theta$, $\tan\theta$, and $\csc\theta$.

• Step 1: Determine the active quadrant.

  • $\cos\theta < 0$ occurs in Quadrants II and III.
  • $\tan\theta > 0$ occurs in Quadrants I and III.
  • Therefore, angle $\theta$ terminates in Quadrant III, where both $x$ and $y$ are negative.

• Step 2: Calculate $\sin\theta$ using $\sin^2\theta + \cos^2\theta = 1$.

  $$\sin^2\theta + \left(-\frac{4}{5}\right)^2 = 1 \implies \sin^2\theta + \frac{16}{25} = 1$$ $$\sin^2\theta = 1 - \frac{16}{25} = \frac{9}{25}$$

  In Quadrant III, sine is negative:

  $$\sin\theta = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$

• Step 3: Calculate $\tan\theta$ and $\csc\theta$.

  $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$$ $$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$$

Worked Problem 2: Proving an Identity Algebraically

Problem: Algebraically verify the identity for all defined values of $\theta$:

$$\frac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta$$

• Step 1: Select the left-hand side (LHS) and multiply by the conjugate.

  $$\text{LHS} = \frac{\cos\theta}{1 - \sin\theta}$$

  Multiply numerator and denominator by $(1 + \sin\theta)$:

  $$\text{LHS} = \frac{\cos\theta(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}$$

• Step 2: Expand denominator using difference of squares.

  $$(1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta$$ $$\text{LHS} = \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta}$$

• Step 3: Apply the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$.

  $$\text{LHS} = \frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta}$$

• Step 4: Cancel the common factor of $\cos\theta$.

  $$\text{LHS} = \frac{1 + \sin\theta}{\cos\theta}$$

• Step 5: Split the fraction and apply reciprocal/quotient definitions.

  $$\text{LHS} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta = \text{RHS} \quad \blacksquare$$

Worked Problem 3: Solving a Quadratic Trigonometric Equation over $[0, 2\pi)$

Problem: Find all solutions to $2\sin^2(x) + \sin(x) - 1 = 0$ on the domain $0 \le x < 2\pi$.

• Step 1: Factor the quadratic expression. Let $u = \sin(x)$: $2u^2 + u - 1 = (2u - 1)(u + 1)$.

  $$(2\sin(x) - 1)(\sin(x) + 1) = 0$$

• Step 2: Set each linear factor to zero.

  $$2\sin(x) - 1 = 0 \implies \sin(x) = \frac{1}{2}$$ $$\sin(x) + 1 = 0 \implies \sin(x) = -1$$

• Step 3: Solve Branch 1: $\sin(x) = \frac{1}{2}$. The reference angle is $x_R = \frac{\pi}{6}$. Sine is positive in Quadrants I and II:

  • Quadrant I: $x = \frac{\pi}{6}$
  • Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

• Step 4: Solve Branch 2: $\sin(x) = -1$. On the unit circle, $y = -1$ occurs at the negative $y$-axis:

  $$x = \frac{3\pi}{2}$$

• Step 5: State the complete solution set.

  $$\left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\right\}$$

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6. Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Dividing by a trigonometric variable and losing solutions. In an equation like $2\sin(x)\cos(x) = \sin(x)$, dividing both sides by $\sin(x)$ eliminates the solutions where $\sin(x) = 0$ ($x = 0, \pi$). Never divide by a variable function. Always subtract to set the equation to zero and factor: $\sin(x)(2\cos(x) - 1) = 0$.
• Pitfall 2: Forgetting to reject invalid roots. If factoring produces $\cos(x) = 2$ or $\sin(x) = -1.5$, you must explicitly state that these branches have no real solution because the range of sine and cosine is $[-1, 1]$.
• Pitfall 3: Treating identity proofs like equations. Performing operations across the equals sign in an identity verification earns a 0 on Regents scoring rubrics. Always transform one side independently.