Question

Given $a_{1} \cos \alpha_{1}+a_{2} \cos \alpha_{2}+\ldots+a_{n} \cos \alpha_{n}=0$ and $a_{1} \cos \left(\alpha_{1}+\theta\right)+a_{2} \cos \left(\alpha_{2}+\theta\right)+\ldots+a_{n} \cos \left(\alpha_{n}+\theta\right)=0$ $(\theta \neq k \pi)$, then the value of $a_{1} \cos \left(\alpha_{1}+\lambda\right)+a_{2} \cos \left(\alpha_{2}+\lambda\right)$ $+\ldots+a_{n} \cos \left(\alpha_{n}+\lambda\right)$ is

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

Answer 1

Answer: (D)

$a_{1} \cos \left(\alpha_{1}+\theta\right)+a_{2} \cos \left(\alpha_{2}+\theta\right)+\ldots+a_{n} \cos \left(\alpha_{n}+\theta\right)=0$
$\Rightarrow\left(a_{1} \cos \alpha_{1}+a_{2} \cos \alpha_{2}+\ldots +a_{n} \cos \alpha_{n}\right) \cos \theta$
$-\left(a_{1} \sin \alpha_{1}+a_{2} \sin \alpha_{2}+\ldots+ a_{n} \sin \alpha_{n}\right) \sin \theta=0$
$\Rightarrow a_{1} \sin \alpha_{1}+a_{2} \sin \alpha_{2}+\ldots +a_{n} \sin \alpha_{n}=0($ since $\sin \theta \neq 0)$
and $a_{1} \cos \alpha_{1}+a_{2} \cos \alpha_{2}+\ldots+ a_{n} \cos \alpha_{n}=0$
Now, $a_{1} \cos \left(\alpha_{1}+\lambda\right)+a_{2} \cos \left(\alpha_{2}+\lambda\right)+\ldots+ a_{n} \cos \left(\alpha_{n}+\lambda\right)$
$=\left(a_{1} \cos \alpha_{1}+a_{2} \cos \alpha_{2}+\ldots+ a_{n} \cos \alpha_{n}\right) \cos \lambda$
$-\left(a_{1} \sin \alpha_{1}+a_{2} \sin \alpha_{2}+\ldots +a_{n} \sin \alpha_{n}\right) \sin \lambda=0$