Question

If $a\sin x+b\cos (x+\theta )+b\cos (x-\theta )=d$ then the minimum value of $|\cos \theta |$ is equal to:

• A. $\frac{1}{2|b|}\sqrt{d^2-a^2}$
• B. $\frac{1}{2|a|}\sqrt{d^2-a^2}$
• C. $\frac{1}{2|d|}\sqrt{d^2-a^2}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2|b|}\sqrt{d^2-a^2}$

$a\sin x+b\cos (x+\theta )+b\cos (x-\theta )=d\Rightarrow a\sin x+b\left[\cos \right(x+\theta )+\cos (x-\theta \left)\right]=d\Rightarrow a\sin x+b(2\cos x+\cos \theta )=d\left(U\sin g\cos \right(a+b)+\cos (a-b)=2\cos a\cos b)\Rightarrow a\sin x+\left(2b\cos \theta \right)\cos x=d$

We know, that $a\sin \theta +b\cos \theta$ lies between $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$

$\therefore$ Min. value of $d=-\sqrt{a^2+{\left(2b\cos \theta \right)}^2}\Rightarrow d\le a^2+{\left(2b\cos \theta \right)}^2$

$\Rightarrow d^2\le a^2+4b^2{\cos }^2\theta \Rightarrow 4b^2{\cos }^2\theta \ge d^2-a^2\Rightarrow {\cos }^2\theta \ge \frac{d^2-a^2}{4b^2}\Rightarrow \cos \theta \ge \frac{\sqrt{d^2-a^2}}{2\left|a\right|}$

Therefore the min value of $\left|\cos \theta \right|=\frac{\sqrt{d^2-a^2}}{2\left|b\right|}$