Problem 40
Question
a. Dertve a formula for \(\cos 3 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\) using Euler's formula. b. Derive the formula \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\) from part (a) and the identity \(\sin ^{2} \theta+\cos ^{2} \theta=1\). (We will have use for this identity in Section 32.)
Step-by-Step Solution
Verified Answer
\(\cos 3\theta = 4 \cos^3 \theta - 3\cos \theta\).
1Step 1: Express Euler's Formula
Euler's formula states that any complex exponential can be expressed as the sum of sine and cosine functions: \( e^{i\theta} = \cos \theta + i \sin \theta \). We can use this to express \(e^{i(3\theta)}\) as \((e^{i\theta})^3\).
2Step 2: Expand \( (e^{i\theta})^3 \) using Binomial Theorem
Calculate \((e^{i\theta})^3 = (\cos \theta + i \sin \theta)^3\) by expanding it using the binomial theorem: \(a^3 + 3a^2b + 3ab^2 + b^3\), where \(a = \cos \theta\) and \(b = i \sin \theta\).
3Step 3: Simplify \((\cos \theta + i \sin \theta)^3\)
Apply the binomial expansion to get: \((\cos \theta)^3 + 3(\cos \theta)^2(i \sin \theta) + 3(\cos \theta)(i \sin \theta)^2 + (i \sin \theta)^3\). Simplify the powers of \(i\): \(i^2 = -1\) and \(i^3 = -i\).
4Step 4: Separate Real and Imaginary Components
After simplifying, collect the real part (cosine terms) and imaginary part (sine terms): \( \cos 3\theta + i \sin 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta + i(3\cos^2 \theta \sin \theta - \sin^3 \theta) \).
5Step 5: Isolate \(\cos 3\theta\)
Identify the real component from the expression obtained in Step 4 as \(\cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta\).
6Step 6: Substitute Identity \(\sin^2 \theta = 1 - \cos^2 \theta\)
Use the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to substitute in \(\cos 3\theta = \cos^3 \theta - 3\cos \theta (1 - \cos^2 \theta)\).
7Step 7: Simplify the Expression
Distribute the \(-3\cos \theta\) and simplify: \( \cos 3\theta = \cos^3 \theta - 3\cos \theta + 3\cos^3 \theta\). Combine like terms to derive: \(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\).
Key Concepts
Trigonometric IdentitiesBinomial TheoremComplex Numbers
Trigonometric Identities
Trigonometric identities are particularly useful tools in mathematics, helping simplify and solve equations involving angles. At the core, these identities relate the angles and sides of triangles, providing a means to express trigonometric functions such as sine, cosine, and tangent in various forms. One of the most fundamental trigonometric identities is the Pythagorean identity:
- \( \sin^2 \theta + \cos^2 \theta = 1 \)
- \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)
- It can also be rewritten in terms of either sine or cosine alone, like \( 2\cos^2 \theta - 1 \) or \( 1 - 2\sin^2 \theta \).
Binomial Theorem
The binomial theorem is a powerful mathematical tool used to expand expressions that are raised to a power, like \( (a + b)^n \). This theorem states:
- \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
- \( \cos^3 \theta + 3\cos^2 \theta(i \sin \theta) + 3\cos \theta(i \sin \theta)^2 + (i \sin \theta)^3 \)
Complex Numbers
Complex numbers are a vast and interesting topic in mathematics, expanding beyond the limitations of real numbers. Represented typically as \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit (\( i^2 = -1 \)), complex numbers allow for the solving of equations that don't have solutions among the reals. Euler's formula itself (
- \( e^{i\theta} = \cos \theta + i \sin \theta \)
Other exercises in this chapter
Problem 38
Derive the fornulas $$ \sin (a+b)=\sin a \cos b+\cos a \sin b $$ and $$ \cos (a+b)=\cos a \cos b-\sin a \sin b $$ by using Euler's formula and computing \(e^{j
View solution Problem 40
a. Dertive a formula for \(\cos 3 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\) using Euler's formula. b. Derive the formula \(\cos 3 \theta=4 \cos ^
View solution Problem 38
Derive the formulas $$ \sin (a+b)=\sin a \cos b+\cos a \sin b $$ and $$ \cos (a+b)=\cos a \cos b-\sin a \sin b $$ by using Euler's formula and computing \(e^{i
View solution