Question

Let $p=a\cos \theta -b\sin \theta$. Then for all real $\theta$

• A. $p>\sqrt{a^2+b^2}$
• B. $p<-\sqrt{a^2+b^2}$
• C. $-\sqrt{a^2+b^2}\le p\le \sqrt{a^2+b^2}$
• D. none of these.

Answer 1

Answer: (C)

$-\sqrt{a^2+b^2}\le p\le \sqrt{a^2+b^2}$

Let, $acos\theta -bsin\theta =p$

Divide and multiply $acos\theta -bsin\theta =p$ by $\sqrt{a^2+b^2}$

i.e. $p=\sqrt{a^2+b^2}(\frac{acos\theta }{\sqrt{a^2+b^2}}-\frac{bsin\theta }{\sqrt{a^2+b^2}})$

Let $\frac{a}{\sqrt{a^2+b^2}}=sinx$ and $\frac{b}{\sqrt{a^2+b^2}}=cosx$

We get,
$p=\sqrt{a^2+b^2}(sinxcos\theta -cosxsin\theta )$

$p=\sqrt{a^2+b^2}sin(x+\theta )$

As $-1\le sinx\le 1$

$\Rightarrow -1\le sin(x+\theta )\le 1$

$\Rightarrow -\sqrt{a^2+b^2}\le \sqrt{a^2+b^2}sin(x+\theta )\le \sqrt{a^2+b^2}$

$\Rightarrow -\sqrt{a^2+b^2}\le p\le \sqrt{a^2+b^2}$