Question

Given $a_1\cos {\alpha }_1+a_2\cos {\alpha }_2+\dots +a_n\cos {\alpha }_n=0$ and $a_1\cos \left({\alpha }_1+\theta \right)+a_2\cos \left({\alpha }_2+\theta \right)+\dots +a_n\cos \left({\alpha }_n+\theta \right)=0$ $(\theta \ne k\pi )$, then the value of $a_1\cos \left({\alpha }_1+\lambda \right)+a_2\cos \left({\alpha }_2+\lambda \right)$ $+\dots +a_n\cos \left({\alpha }_n+\lambda \right)$ is

• A. $\theta -\lambda$
• B. $\theta +\lambda$
• C. $\lambda$
• D. $0$

Answer 1

Answer: (D)

$a_1\cos \left({\alpha }_1+\theta \right)+a_2\cos \left({\alpha }_2+\theta \right)+\dots +a_n\cos \left({\alpha }_n+\theta \right)=0$
$\Rightarrow \left(a_1\cos {\alpha }_1+a_2\cos {\alpha }_2+\dots +a_n\cos {\alpha }_n\right)\cos \theta$
$-\left(a_1\sin {\alpha }_1+a_2\sin {\alpha }_2+\dots +a_n\sin {\alpha }_n\right)\sin \theta =0$
$\Rightarrow a_1\sin {\alpha }_1+a_2\sin {\alpha }_2+\dots +a_n\sin {\alpha }_n=0($ since $\sin \theta \ne 0)$
and $a_1\cos {\alpha }_1+a_2\cos {\alpha }_2+\dots +a_n\cos {\alpha }_n=0$
Now, $a_1\cos \left({\alpha }_1+\lambda \right)+a_2\cos \left({\alpha }_2+\lambda \right)+\dots +a_n\cos \left({\alpha }_n+\lambda \right)$
$=\left(a_1\cos {\alpha }_1+a_2\cos {\alpha }_2+\dots +a_n\cos {\alpha }_n\right)\cos \lambda$
$-\left(a_1\sin {\alpha }_1+a_2\sin {\alpha }_2+\dots +a_n\sin {\alpha }_n\right)\sin \lambda =0$