Math Problem Statement

(1 + sin theta + i * cos theta) ^ n + (1 + sin theta - i * cos theta) ^ n = 2 ^ (n + 1) * cos^n (pi/4 - theta/2) * cos((npi)/4 - (ntheta)/2)

Solution

We are tasked with proving the following equation:

(1+sinθ+icosθ)n+(1+sinθicosθ)n=2n+1cosn(π4θ2)cos(nπ4nθ2)(1 + \sin \theta + i \cos \theta)^n + (1 + \sin \theta - i \cos \theta)^n = 2^{n+1} \cos^n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \cos \left( \frac{n\pi}{4} - \frac{n\theta}{2} \right)

Step 1: Complex Form Simplification

The terms on the left-hand side involve complex numbers of the form z1=1+sinθ+icosθz_1 = 1 + \sin \theta + i \cos \theta and z2=1+sinθicosθz_2 = 1 + \sin \theta - i \cos \theta. These can be written as:

z1=(1+sinθ)+icosθz_1 = (1 + \sin \theta) + i \cos \theta z2=(1+sinθ)icosθz_2 = (1 + \sin \theta) - i \cos \theta

Step 2: Converting to Exponential Form

Recognize that each complex number can be rewritten in polar form. For z1z_1 and z2z_2, use the fact that 1+sinθ1 + \sin \theta is the real part and cosθ\cos \theta is the imaginary part.

z1=(1+sinθ)2+cos2θeiϕz_1 = \sqrt{(1 + \sin \theta)^2 + \cos^2 \theta} e^{i \phi}

Where the magnitude z1=(1+sinθ)2+cos2θ=1+2sinθ+sin2θ+cos2θ=2+2sinθ=2cos(π4θ2)|z_1| = \sqrt{(1 + \sin \theta)^2 + \cos^2 \theta} = \sqrt{1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta} = \sqrt{2 + 2\sin \theta} = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right).

Thus,

z1=2cos(π4θ2)ei(π4θ2)z_1 = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) e^{i \left( \frac{\pi}{4} - \frac{\theta}{2} \right)} and z2=2cos(π4θ2)ei(π4θ2).z_2 = 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) e^{-i \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}.

Step 3: Raising to Power nn

Now, raise z1z_1 and z2z_2 to the power nn:

z1n=[2cos(π4θ2)]nein(π4θ2)z_1^n = \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n e^{in \left( \frac{\pi}{4} - \frac{\theta}{2} \right)} and z2n=[2cos(π4θ2)]nein(π4θ2).z_2^n = \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n e^{-in \left( \frac{\pi}{4} - \frac{\theta}{2} \right)}.

Step 4: Summing z1nz_1^n and z2nz_2^n

Now, we sum z1nz_1^n and z2nz_2^n:

z1n+z2n=2[2cos(π4θ2)]ncos(n(π4θ2))z_1^n + z_2^n = 2 \left[ 2 \cos \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right]^n \cos \left( n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right)

=2n+1cosn(π4θ2)cos(nπ4nθ2)= 2^{n+1} \cos^n \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \cos \left( \frac{n\pi}{4} - \frac{n\theta}{2} \right)

This matches the right-hand side of the equation we set out to prove.

Thus, the equation is verified.


Would you like more details or clarification on any step?

Here are 5 questions to consider based on this explanation:

  1. What is the significance of using polar form for complex numbers in this proof?
  2. How would the proof change if θ\theta were negative?
  3. Can this technique of summing powers of complex numbers be applied to other trigonometric identities?
  4. How can the relationship between the magnitude and argument of complex numbers help in solving other trigonometric equations?
  5. What role does Euler's formula play in simplifying trigonometric expressions involving complex numbers?

Tip: Understanding polar and exponential forms of complex numbers is key to simplifying expressions involving powers and products of complex numbers.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Coordinates
Trigonometric Identities
Exponentiation

Formulas

(1 + sin θ + i * cos θ)^n + (1 + sin θ - i * cos θ)^n
2^(n+1) * cos^n(π/4 - θ/2) * cos((nπ)/4 - (nθ)/2)
Polar form of complex numbers: z = r e^(iθ)
Euler's formula: e^(iθ) = cos(θ) + i sin(θ)

Theorems

Euler's Formula
De Moivre's Theorem

Suitable Grade Level

College level or advanced high school