📐 Level 2 - Topic 10: Double Angle Identities - Doubling the Angle Power 2️⃣🔗

1) Unleashing Double Angle Identities - Doubling the Angle 2️⃣

Welcome to the realm of double angle identities! These are a special set of trigonometric identities derived from the sum identities. They provide expressions for trigonometric functions of \( 2\theta \) in terms of trigonometric functions of \( \theta \). Double angle identities are powerful tools for simplifying expressions, solving equations, and are fundamental in calculus and various applications in science and engineering.

What are Double Angle Identities?

  • They are formulas that express \( \sin(2\theta) \), \( \cos(2\theta) \), and \( \tan(2\theta) \) using trigonometric functions of \( \theta \).
  • They are derived directly from the sum identities by setting \( B = A \).

Why are Double Angle Identities Important?

  • Simplification: Simplify expressions involving trigonometric functions of double angles.
  • Solving Equations: Transform equations involving \( 2\theta \) and \( \theta \) into equations with a single angle variable.
  • Integration and Differentiation: Essential in calculus when dealing with trigonometric functions.
  • Further Identities: Serve as a stepping stone for deriving other identities like half-angle and power-reducing identities.

In this topic, we will derive and explore the double angle identities for sine, cosine, and tangent. We will also learn how to apply them in simplification, verification, and solving trigonometric equations. Let's double the angle and unlock new trigonometric capabilities! 2️⃣🔗

2) Double Angle Identities for Sine - Sine of Double Angle 🎶

Let's begin with the double angle identity for the sine function.

Definition: Double Angle Identity for Sine

The double angle identity for sine is:

\( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)

Derivation from Sum Identity:
  • Start with the sine sum identity: \( \sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) \).
  • To get the double angle identity, set \( B = A = \theta \).
  • Substitute \( A = \theta \) and \( B = \theta \) into the sum identity:

    \( \sin(\theta + \theta) = \sin(\theta)\cos(\theta) + \cos(\theta)\sin(\theta) \)

    \( \sin(2\theta) = \sin(\theta)\cos(\theta) + \sin(\theta)\cos(\theta) \)

    \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)

Example 1: Using Sine Double Angle Identity

If \( \sin(\theta) = \frac{3}{5} \) and \( \theta \) is in quadrant II, find \( \sin(2\theta) \).

  1. Need \( \cos(\theta) \): Since \( \sin^2(\theta) + \cos^2(\theta) = 1 \), \( \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \). So, \( \cos(\theta) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5} \).
  2. Determine sign of \( \cos(\theta) \): In quadrant II, cosine is negative. So, \( \cos(\theta) = -\frac{4}{5} \).
  3. Apply sine double angle identity: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).

    \( \sin(2\theta) = 2 \cdot \left(\frac{3}{5}\right) \cdot \left(-\frac{4}{5}\right) \)

  4. Simplify:

    \( \sin(2\theta) = -\frac{2 \cdot 3 \cdot 4}{5 \cdot 5} = -\frac{24}{25} \)

Solution: \( \sin(2\theta) = -\frac{24}{25} \).

3) Double Angle Identities for Cosine - Cosine of Double Angle 🎵

Next, let's derive and explore the double angle identities for the cosine function. Cosine has multiple forms for its double angle identity, which provide flexibility in different situations.

Definition: Double Angle Identities for Cosine

There are three common forms for the double angle identity for cosine:

Form 1 (in terms of sine and cosine):

\( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \)

Form 2 (in terms of cosine only):

\( \cos(2\theta) = 2\cos^2(\theta) - 1 \)

Form 3 (in terms of sine only):

\( \cos(2\theta) = 1 - 2\sin^2(\theta) \)

Derivation from Cosine Sum Identity:
  • Start with the cosine sum identity: \( \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \).
  • Set \( B = A = \theta \) to derive the first form.

    \( \cos(\theta + \theta) = \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta) \)

    \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \)

  • To get Form 2 (in terms of cosine only), use Pythagorean identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \) in Form 1:

    \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1 - \cos^2(\theta)) \)

    \( \cos(2\theta) = \cos^2(\theta) - 1 + \cos^2(\theta) \)

    \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)

  • To get Form 3 (in terms of sine only), use Pythagorean identity \( \cos^2(\theta) = 1 - \sin^2(\theta) \) in Form 1:

    \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = (1 - \sin^2(\theta)) - \sin^2(\theta) \)

    \( \cos(2\theta) = 1 - \sin^2(\theta) - \sin^2(\theta) \)

    \( \cos(2\theta) = 1 - 2\sin^2(\theta) \)

Example 2: Using Cosine Double Angle Identity (Form 1)

If \( \sin(\theta) = \frac{3}{5} \) and \( \theta \) is in quadrant II, find \( \cos(2\theta) \) using Form 1: \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \).

  1. We already know from Example 1: \( \sin(\theta) = \frac{3}{5} \) and \( \cos(\theta) = -\frac{4}{5} \).
  2. Apply cosine double angle identity (Form 1): \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \).

    \( \cos(2\theta) = \left(-\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 \)

  3. Simplify:

    \( \cos(2\theta) = \frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25} \)

Solution: \( \cos(2\theta) = \frac{7}{25} \).

Example 3: Using Cosine Double Angle Identity (Form 2)

Given \( \cos(\theta) = -\frac{4}{5} \), find \( \cos(2\theta) \) using Form 2: \( \cos(2\theta) = 2\cos^2(\theta) - 1 \).

  1. Use Form 2 directly with \( \cos(\theta) = -\frac{4}{5} \): \( \cos(2\theta) = 2\cos^2(\theta) - 1 \).

    \( \cos(2\theta) = 2 \cdot \left(-\frac{4}{5}\right)^2 - 1 \)

  2. Simplify:

    \( \cos(2\theta) = 2 \cdot \left(\frac{16}{25}\right) - 1 = \frac{32}{25} - 1 = \frac{32 - 25}{25} = \frac{7}{25} \)

Solution: \( \cos(2\theta) = \frac{7}{25} \) (Same result as Example 2, as expected).

Example 4: Using Cosine Double Angle Identity (Form 3)

Given \( \sin(\theta) = \frac{3}{5} \), find \( \cos(2\theta) \) using Form 3: \( \cos(2\theta) = 1 - 2\sin^2(\theta) \).

  1. Use Form 3 directly with \( \sin(\theta) = \frac{3}{5} \): \( \cos(2\theta) = 1 - 2\sin^2(\theta) \).

    \( \cos(2\theta) = 1 - 2 \cdot \left(\frac{3}{5}\right)^2 \)

  2. Simplify:

    \( \cos(2\theta) = 1 - 2 \cdot \left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{25 - 18}{25} = \frac{7}{25} \)

Solution: \( \cos(2\theta) = \frac{7}{25} \) (Again, same result, showing consistency across forms).

4) Double Angle Identity for Tangent - Tangent of Double Angle 🥁

Finally, let's derive and examine the double angle identity for the tangent function.

Definition: Double Angle Identity for Tangent

The double angle identity for tangent is:

\( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \)

Derivation from Tangent Sum Identity:
  • Start with the tangent sum identity: \( \tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \).
  • Set \( B = A = \theta \).
  • Substitute \( A = \theta \) and \( B = \theta \) into the sum identity:

    \( \tan(\theta + \theta) = \frac{\tan(\theta) + \tan(\theta)}{1 - \tan(\theta)\tan(\theta)} \)

    \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \)

Example 5: Using Tangent Double Angle Identity

If \( \sin(\theta) = \frac{3}{5} \) and \( \theta \) is in quadrant II, find \( \tan(2\theta) \).

  1. We know from Example 1: \( \sin(\theta) = \frac{3}{5} \) and \( \cos(\theta) = -\frac{4}{5} \).
  2. Find \( \tan(\theta) \): \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{3/5}{-4/5} = -\frac{3}{4} \).
  3. Apply tangent double angle identity: \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \).

    \( \tan(2\theta) = \frac{2 \cdot \left(-\frac{3}{4}\right)}{1 - \left(-\frac{3}{4}\right)^2} \)

  4. Simplify:

    \( \tan(2\theta) = \frac{-\frac{3}{2}}{1 - \frac{9}{16}} = \frac{-\frac{3}{2}}{\frac{16 - 9}{16}} = \frac{-\frac{3}{2}}{\frac{7}{16}} = -\frac{3}{2} \cdot \frac{16}{7} = -\frac{3 \cdot 8}{7} = -\frac{24}{7} \)

Solution: \( \tan(2\theta) = -\frac{24}{7} \).

Check consistency: From Examples 1 and 2/3/4, \( \sin(2\theta) = -\frac{24}{25} \) and \( \cos(2\theta) = \frac{7}{25} \). Then \( \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{-24/25}{7/25} = -\frac{24}{7} \). Results are consistent! 🎉

5) Verifying Identities using Double Angle Identities 🚀

Example 6: Verifying Identity using Cosine Double Angle Identity

Verify the identity: \( \frac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan(\theta) \).

  1. Start with the left side (LS): \( \text{LS} = \frac{\sin(2\theta)}{1 + \cos(2\theta)} \).
  2. Apply double angle identities: Use \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \) and \( \cos(2\theta) = 2\cos^2(\theta) - 1 \) (Form 2 of cosine double angle).

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

  3. Simplify the denominator: \( 1 + (2\cos^2(\theta) - 1) = 2\cos^2(\theta) \).

    \( \text{LS} = \frac{2\sin(\theta)\cos(\theta)}{2\cos^2(\theta)} \)

  4. Simplify by cancelling common factors: Cancel \( 2\cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)).

    \( \text{LS} = \frac{\sin(\theta)}{\cos(\theta)} \)

  5. Use quotient identity for \( \tan(\theta) \): \( \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \). So, \( \text{LS} = \tan(\theta) \).
  6. Compare with the right side (RS): \( \text{RS} = \tan(\theta) \).
  7. Conclusion: Since \( \text{LS} = \text{RS} \), the identity \( \frac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan(\theta) \) is verified.

Example 7: Verifying Identity using Cosine Double Angle Identity (Form 3)

Verify the identity: \( \frac{1 - \cos(2\theta)}{\sin(2\theta)} = \tan(\theta) \).

  1. Start with the left side (LS): \( \text{LS} = \frac{1 - \cos(2\theta)}{\sin(2\theta)} \).
  2. Apply double angle identities: Use \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \) and \( \cos(2\theta) = 1 - 2\sin^2(\theta) \) (Form 3 of cosine double angle).

    \( \text{LS} = \frac{1 - (1 - 2\sin^2(\theta))}{\sin(2\theta)} = \frac{1 - 1 + 2\sin^2(\theta)}{\sin(2\theta)} = \frac{2\sin^2(\theta)}{\sin(2\theta)} \)

  3. Use sine double angle identity in the denominator: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).

    \( \text{LS} = \frac{2\sin^2(\theta)}{2\sin(\theta)\cos(\theta)} \)

  4. Simplify by cancelling common factors: Cancel \( 2\sin(\theta) \) (assuming \( \sin(\theta) \neq 0 \)).

    \( \text{LS} = \frac{\sin(\theta)}{\cos(\theta)} \)

  5. Use quotient identity for \( \tan(\theta) \): \( \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \). So, \( \text{LS} = \tan(\theta) \).
  6. Compare with the right side (RS): \( \text{RS} = \tan(\theta) \).
  7. Conclusion: Since \( \text{LS} = \text{RS} \), the identity \( \frac{1 - \cos(2\theta)}{\sin(2\theta)} = \tan(\theta) \) is verified.

6) Solving Equations using Double Angle Identities 🚀

Example 8: Solving Equation using Sine Double Angle Identity

Solve the equation \( \sin(2x) = \cos(x) \) for principal solutions in \( [0, 2\pi) \).

  1. Apply sine double angle identity: Replace \( \sin(2x) \) with \( 2\sin(x)\cos(x) \). Equation becomes \( 2\sin(x)\cos(x) = \cos(x) \).
  2. Rearrange and Factor: \( 2\sin(x)\cos(x) - \cos(x) = 0 \Rightarrow \cos(x)(2\sin(x) - 1) = 0 \).
  3. Set each factor to zero: \( \cos(x) = 0 \) or \( 2\sin(x) - 1 = 0 \Rightarrow \sin(x) = \frac{1}{2} \).
  4. Solve \( \cos(x) = 0 \): Principal solutions in \( [0, 2\pi) \) are \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).
  5. Solve \( \sin(x) = \frac{1}{2} \): Principal solutions in \( [0, 2\pi) \) are \( x = \frac{\pi}{6}, \frac{5\pi}{6} \).
  6. Combine all principal solutions: \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} \).

Solution: Principal solutions are \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} \).

Example 9: Solving Equation using Cosine Double Angle Identity

Solve the equation \( \cos(2x) + \cos(x) = 0 \) for principal solutions in \( [0, 2\pi) \).

  1. Apply cosine double angle identity: Use \( \cos(2x) = 2\cos^2(x) - 1 \) (Form 2 of cosine double angle). Equation becomes \( (2\cos^2(x) - 1) + \cos(x) = 0 \Rightarrow 2\cos^2(x) + \cos(x) - 1 = 0 \).
  2. Quadratic equation in \( \cos(x) \): Let \( y = \cos(x) \). Equation becomes \( 2y^2 + y - 1 = 0 \).
  3. Solve quadratic equation: Factor \( (2y - 1)(y + 1) = 0 \). Solutions are \( y = \frac{1}{2} \) or \( y = -1 \).
  4. Back-substitute \( y = \cos(x) \): \( \cos(x) = \frac{1}{2} \) or \( \cos(x) = -1 \).
  5. Solve \( \cos(x) = \frac{1}{2} \): Principal solutions in \( [0, 2\pi) \) are \( x = \frac{\pi}{3}, \frac{5\pi}{3} \).
  6. Solve \( \cos(x) = -1 \): Principal solution in \( [0, 2\pi) \) is \( x = \pi \).
  7. Combine all principal solutions: \( x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).

Solution: Principal solutions are \( x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).

7) Practice Questions 🎯

7.1 Fundamental – Using Double Angle Identities for Calculation and Verification

1. If \( \cos(\theta) = \frac{12}{13} \) and \( \theta \) is in quadrant I, find \( \sin(2\theta) \), \( \cos(2\theta) \), and \( \tan(2\theta) \).

2. If \( \tan(\theta) = -\frac{4}{3} \) and \( \theta \) is in quadrant IV, find \( \sin(2\theta) \), \( \cos(2\theta) \), and \( \tan(2\theta) \).

3. Express \( \sin(3\theta) \) in terms of \( \sin(\theta) \) and \( \cos(\theta) \) using sum and double angle identities (Hint: \( \sin(3\theta) = \sin(2\theta + \theta) \)).

4. Express \( \cos(3\theta) \) in terms of \( \cos(\theta) \) only (Hint: \( \cos(3\theta) = \cos(2\theta + \theta) \) and use cosine double angle Form 2).

5. Verify the identity: \( \cos(2\theta) = 1 - \tan^2(\theta) \) / \( 1 + \tan^2(\theta) \) (Express everything in terms of sin and cos).

6. Verify the identity: \( \frac{\sin(2\theta)}{\sin(\theta)} = 2\cos(\theta) \).

7. Verify the identity: \( \frac{1 + \cos(2\theta)}{2} = \cos^2(\theta) \) (Power-reducing identity).

8. Verify the identity: \( \frac{1 - \cos(2\theta)}{2} = \sin^2(\theta) \) (Power-reducing identity).

9. Simplify: \( 2\sin(5x)\cos(5x) \).

10. Simplify: \( \cos^2(3x) - \sin^2(3x) \).

11. Simplify: \( 1 - 2\sin^2(4x) \).

12. Simplify: \( \frac{2\tan(x/2)}{1 - \tan^2(x/2)} \).

7.2 Challenging – Advanced Verification and Equation Solving 💪🚀

1. Verify: \( \frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \sin(2\theta) \) (Express everything in terms of sin and cos).

2. Verify: \( \cot(2\theta) = \frac{\cot^2(\theta) - 1}{2\cot(\theta)} \) (Express cot in terms of tan or sin/cos).

3. Verify: \( \tan(\theta) = \frac{\sin(2\theta)}{1 + \cos(2\theta)} \) (Already verified in Example 6, now try to verify starting from the other side).

4. Solve the equation: \( \sin(2x) + \sin(x) = 0 \) for principal solutions in \( [0, 2\pi) \).

5. Solve the equation: \( \cos(2x) = \sin(x) \) for principal solutions in \( [0^\circ, 360^\circ) \).

6. Solve the equation: \( \tan(2x) = \tan(x) \) for general solutions in radians.

7. (Conceptual) Explain why using different forms of \( \cos(2\theta) \) identity can be beneficial in different contexts. Give examples.

8. (Challenging Verification) Verify: \( \frac{1 - \cos(4\theta)}{\tan(\theta) - \cot(\theta)} = -2\sin^3(2\theta) \) (Use double angle identities and simplify).

8) Summary - Double Angle Identities - Power of Two 🎉

  • Double Angle Identities: Formulas for \( \sin(2\theta) \), \( \cos(2\theta) \), \( \tan(2\theta) \) in terms of functions of \( \theta \).
  • Sine Double Angle: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).
  • Cosine Double Angle: Three forms: \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \), \( \cos(2\theta) = 2\cos^2(\theta) - 1 \), \( \cos(2\theta) = 1 - 2\sin^2(\theta) \). Choose the form based on the problem.
  • Tangent Double Angle: \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \).
  • Derivation: Derived from sum identities by setting \( B = A = \theta \).
  • Applications: Simplifying expressions, verifying identities, solving equations, calculus.
  • Techniques: Recognize double angle forms, apply identities to simplify or transform, algebraic manipulation, factor quadratic forms.

Excellent! You've now mastered the double angle identities, another powerful set of tools in trigonometry. These identities allow you to relate trigonometric functions of an angle to those of its double, enabling you to solve a broader range of problems and further expand your trigonometric capabilities. As you continue to the next topics, you will see how these identities are used to derive even more advanced identities and solve more complex trigonometric problems in mathematics and beyond. Keep practicing, and you will master the art of manipulating angles! 2️⃣🔗💪📐🌟

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