Question

The real part of \( \frac{\left( \cos a + i \sin a \right)^6}{\left( \sin b + i \cos b \right)^8} \) is:

• A. \( \sin (6a - 8b) \)
• B. \( \cos (6a - 8b) \)
• C. \( \sin (6a + 8b) \)
• D. \( \cos (6a + 8b) \)

Hint:

For simplifying complex expressions with trigonometric functions, always use De Moivre's theorem for powers and angle addition formulas to break down the terms into simpler components.

Answer 1

The Correct Option is D

To solve for the real part of the given expression, we first simplify it: \( \frac{\left( \cos a + i \sin a \right)^6}{\left( \sin b + i \cos b \right)^8} \) By applying De Moivre's theorem, we know that

: \( \left( \cos a + i \sin a \right)^6 = \cos (6a) + i \sin (6a) \) \( \left( \sin b + i \cos b \right)^8 = \left( \cos b + i \sin b \right)^8 = \cos (8b) + i \sin (8b) \) 

Thus, the given expression becomes:

 \( \frac{\cos (6a) + i \sin (6a)}{\cos (8b) + i \sin (8b)} \) 

Now, to find the real part of this complex number, we multiply both the numerator and denominator by the conjugate of the denominator:

 \( \frac{(\cos (6a) + i \sin (6a)) (\cos (8b) - i \sin (8b))}{(\cos (8b) + i \sin (8b)) (\cos (8b) - i \sin (8b))} \) The denominator simplifies to: \( \cos^2(8b) + \sin^2(8b) = 1 \) 

Now, for the numerator: 

\( (\cos (6a) + i \sin (6a)) (\cos (8b) - i \sin (8b)) = \cos(6a)\cos(8b) + \sin(6a)\sin(8b) + i \left( \sin(6a)\cos(8b) - \cos(6a)\sin(8b) \right) \) 

Using the angle addition formulas for cosine and sine: 

\( = \cos(6a + 8b) + i \sin(6a + 8b) \) Thus, the real part of the given expression is: \( \cos(6a + 8b) \) Therefore, the correct answer is \( \boxed{\cos(6a + 8b)} \), corresponding to option (4)