Question

For m, $n\in N$,
Let A(n) = $16\left({\cos }^n\theta -{\sin }^n\theta \right)$ and B(m) = $16\left({\cos }^m\theta +{\sin }^m\theta \right)$ and ${\log }_{10}2=0.3010$
If A(6) = $a\cos 2\theta +b\cos 6\theta$, then value of (a+b) is equal to

• A. $12$
• B. $13$
• C. $14$
• D. $16$

Answer 1

The correct option is D $16$

$A\left(n\right)=16(\cos {}^n\theta -\sin {}^n\theta )B\left(m\right)=16(\cos {}^m\theta +\sin {}^m\theta ){\log }_{10}2=0.3010A\left(6\right)=a\cos 2\theta +b\cos 6\theta \therefore 16(\cos {}^6\theta -\sin {}^6\theta )=a\cos 2\theta +b\cos 6\theta \therefore LHS=16(\cos {}^6\theta -\sin {}^6\theta )\cos x=\frac{e^{ix}+e^{-ix}}{2}\sin x=\frac{e^{ix}-e^{-ix}}{2i}\therefore \sin {}^{6}x=\frac{{(e^{ix}-e^{-ix})}^6}{-64}\therefore \cos {}^{6}x=\frac{{(e^{ix}+e^{-ix})}^6}{64}\therefore LHS=16\left[\frac{1}{64}\right(-e^{6ix}+6e^{4ix}-15e^{2ix}+20-15e^{-2ix}+6e^{-4ix}-e^{-6ix}-e^{-6ix}-6e^{4ix}-15e^{2ix}-20-15e^{-2ix}-6e^{-4ix}-e^{-6ix}\left)\right]=16\left[\frac{1}{32}\right({-e}^{6ix}-e^{-6ix}-15e^{2ix}-15e^{-2ix}\left)\right]=16\left[\frac{-1}{16}\right](\cos 6x+15\cos 2x)|a+b|=|15+1|=16$