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. $\frac{1}{2|a|}\sqrt{d^2-b^2}$
• C. $\frac{1}{2}|b|\sqrt{d^2-a^2}$
• D. $\frac{1}{2}|a|\sqrt{d^2-b^2}$

Answer 1

The correct option is A

$\frac{1}{2|b|}\sqrt{d^2-a^2}$

$a\sin x+b\cos (x+\theta )+b\cos (x-\theta )=d$

$\Rightarrow a\sin x+2b\cos x\cos \theta =d$

$\Rightarrow |d|\le \sqrt{a^2+4b^2{\cos }^2\theta }$

$\Rightarrow \frac{d^2-a^2}{4b^2}\le {\cos }^2\theta$

$|\cos \theta |\ge \frac{\sqrt{d^2-a^2}}{2|b|}$