Question

If $z=\cos 2\theta +i\sin 2\theta$ then which is correct

• A. $\sum _{r=0}^n{C}_r\cos 2r\theta =2^n{\cos }^n\theta \cos n\theta$
• B. $\sum _{r=1}^n{C}_r\cos 2r\theta =2^n{\sin }^n\theta \cos n\theta$
• C. ${\sum }_{r=0}^n{C}_r\sin 2r\theta =2^n{\cos }^n\theta \sin n\theta$
• D. $\sum _{r=0}^n{C}_r\sin 2r\theta =2^n{\sin }^n\theta \sin n\theta$

Answer 1

Answer: (A), (C)

The correct options are
A $\sum _{r=0}^n{C}_r\cos 2r\theta =2^n{\cos }^n\theta \cos n\theta$
C ${\sum }_{r=0}^n{C}_r\sin 2r\theta =2^n{\cos }^n\theta \sin n\theta$

By Binomial Theorem
${(1+z)}^n=C_0+C_{1}z+C_2{z}^2+C_3{z}^3+....+C_n{z}^n={\sum }_{r=0}^n{C}_r{z}^r$ ...(1)
Substituting $z=\cos 2\theta +i\sin 2\theta$ in eq. (1), we get
${(1+\cos 2\theta +i\sin 2\theta )}^n={\sum }_{r=0}^n{C}_r{(\cos 2\theta +i\sin 2\theta )}^r$
$\Rightarrow {\sum }_{r=0}^n{C}_r(\cos 2r\theta +i\sin 2r\theta )$ ...{De Moivre's Theorem}
$\Rightarrow {\left[2\cos \theta (\cos \theta +i\sin \theta )\right]}^n={\sum }_{r=0}^n{C}_r\cos 2r\theta +i{\sum }_{r=0}^n{C}_r\sin 2r\theta$
$\Rightarrow {\sum }_{r=0}^n{C}_r\cos 2r\theta +i{\sum }_{r=0}^n{C}_r\sin 2r\theta =2^n{\cos }^n\theta (\cos n\theta +i\sin n\theta )$ ...{De Moivre's Theorem}

$\Rightarrow {\sum }_{r=0}^n{C}_r\cos 2r\theta +i\left({\sum }_{r=0}^n{C}_r\sin 2r\theta \right)=2^n{\cos }^n\theta \cos n\theta +i\left(2^n{\cos }^n\theta \sin n\theta \right)$
On comparing real and Imaginary parts, we get
${\sum }_{r=0}^n{C}_r\cos 2r\theta =2^n{\cos }^n\theta \cos n\theta \&{\sum }_{r=0}^n{C}_r\sin 2r\theta =2^n{\cos }^n\theta \sin n\theta$
Hence, option 'A' and 'C' are correct.