Question

Which one of the following statements is correct?
Given, \(\displaystyle \binom{n}{m} = \frac{n!}{m!(n-m)!}\) is the binomial coefficient.

• A. \(\cos n\theta = \cos^n \theta - \binom{n}{2} \cos^{n-2}\theta \sin^2\theta + \binom{n}{4} \cos^{n-4}\theta \sin^4\theta - \dots\)
• B. \(\sin n\theta = \binom{n}{1}\cos^{n-1}\theta \sin\theta + \binom{n}{3}\cos^{n-3}\theta \sin^3\theta + \dots\)
• C. \(\cos n\theta = \cos^n \theta + \binom{n}{2} \cos^{n-2}\theta \sin^2\theta + \binom{n}{4} \cos^{n-4}\theta \sin^4\theta + \dots\)
• D. \(\sin n\theta = \cos^n \theta - \binom{n}{2} \cos^{n-2}\theta \sin^2\theta + \binom{n}{4} \cos^{n-4}\theta \sin^4\theta - \dots\)

Hint:

Use De Moivre's theorem to derive expressions for multiple-angle trigonometric identities in terms of powers of sine and cosine.

Answer 1

The Correct Option is A

Step 1: Using De Moivre's theorem. 
According to De Moivre's theorem: \[ (\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta \]

Step 2: Expand using binomial theorem. 
\[ (\cos \theta + i \sin \theta)^n = \sum_{k=0}^{n} \binom{n}{k} (\cos \theta)^{n-k} (i \sin \theta)^k \]

Step 3: Separate real and imaginary parts. 
- Real part gives \(\cos n\theta = \cos^n \theta - \binom{n}{2} \cos^{n-2}\theta \sin^2\theta + \binom{n}{4} \cos^{n-4}\theta \sin^4\theta - \dots\) 

- Imaginary part gives \(\sin n\theta = \binom{n}{1} \cos^{n-1}\theta \sin\theta - \binom{n}{3} \cos^{n-3}\theta \sin^3\theta + \dots\)

Step 4: Conclusion. 
Hence, option (A) correctly represents the real part, which gives the expression for \(\cos n\theta.\)