What is the Cosine Triple Angle Formula?

The cosine triple angle formula is a trigonometric identity that expresses the cosine of three times an angle in terms of the cosine of the original angle. This formula is particularly useful in simplifying expressions and solving equations involving trigonometric functions.

The Formula

The cosine triple angle formula is given by:

$cos(3theta) = 4cos^3(theta) – 3cos(theta)$

Derivation of the Formula

To understand where this formula comes from, let’s derive it step-by-step using basic trigonometric identities.

  1. Use the Angle Addition Formula
    First, recall the angle addition formula for cosine:

    $cos(a + b) = cos(a)cos(b) – sin(a)sin(b)$

    We can use this to express $cos(3theta)$ as $cos(2theta + theta)$:

    $cos(3theta) = cos(2theta + theta)$

  1. Apply the Angle Addition Formula
    Using the angle addition formula, we get:

    $cos(3theta) = cos(2theta)cos(theta) – sin(2theta)sin(theta)$

  1. Use Double Angle Formulas
    Next, we need the double angle formulas for cosine and sine:

    $cos(2theta) = 2cos^2(theta) – 1$

    $sin(2theta) = 2sin(theta)cos(theta)$

    Substitute these into our expression:

    $cos(3theta) = (2cos^2(theta) – 1)cos(theta) – (2sin(theta)cos(theta))sin(theta)$

  1. Simplify the Expression
    Now, simplify the expression step-by-step:

    $cos(3theta) = 2cos^3(theta) – cos(theta) – 2sin^2(theta)cos(theta)$

    Using the Pythagorean identity $sin^2(theta) = 1 – cos^2(theta)$, we can replace $sin^2(theta)$:

    $cos(3theta) = 2cos^3(theta) – cos(theta) – 2(1 – cos^2(theta))cos(theta)$

    Distribute $cos(theta)$ through the parentheses:

    $cos(3theta) = 2cos^3(theta) – cos(theta) – 2cos(theta) + 2cos^3(theta)$

    Combine like terms:

    $cos(3theta) = 4cos^3(theta) – 3cos(theta)$

    And there you have it—the cosine triple angle formula!

Examples and Applications

Example 1: Simplifying an Expression

Let’s use the cosine triple angle formula to simplify $cos(3theta)$ when $cos(theta) = frac{1}{2}$

Substitute $cos(theta) = frac{1}{2}$ into the formula:

$cos(3theta) = 4left(frac{1}{2}right)^3 – 3left(frac{1}{2}right)$

Calculate the powers and products:

$cos(3theta) = 4left(frac{1}{8}right) – 3left(frac{1}{2}right)$

$cos(3theta) = frac{4}{8} – frac{3}{2}$

$cos(3theta) = frac{1}{2} – frac{3}{2}$

$cos(3theta) = -1$

Example 2: Solving an Equation

Suppose we need to solve the equation $cos(3theta) = frac{1}{2}$

Using the formula, we get:

$4cos^3(theta) – 3cos(theta) = frac{1}{2}$

Let $x = cos(theta)$. Then the equation becomes a cubic equation:

$4x^3 – 3x – frac{1}{2} = 0$

This equation can be solved using numerical methods or factoring techniques, depending on the context.

Conclusion

Understanding the cosine triple angle formula and its derivation can be incredibly useful in various mathematical and engineering applications. It simplifies complex trigonometric expressions and helps solve equations involving multiple angles.

By practicing with examples, you can become more comfortable using this formula and applying it to different problems.

Citations

  1. 1. Khan Academy – Trigonometric Identities
  2. 2. Math is Fun – Trigonometric Identities
  3. 3. Paul’s Online Math Notes – Trig Formulas

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ