Question

If $z=\alpha \sin \theta +\beta \cos \theta$, where $\alpha$ and $\beta$ are positive constants. The maximum value of $z$ is

• A. $\alpha$
• B. $\beta$
• C. $\sqrt{{\alpha }^2+{\beta }^2}$
• D. $\alpha +\beta$

Answer 1

The correct option is C $\sqrt{{\alpha }^2+{\beta }^2}$

given,

$z=\alpha \sin \theta +\beta \cos \theta$

$z=\sqrt{{\alpha }^2+{\beta }^2}[\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}\sin \theta +\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}\cos \theta ]$

let,

$\cos \varphi =\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}$

$\Rightarrow \sin \varphi =\sqrt{1-{\cos }^2\varphi }=\sqrt{1-\frac{{\alpha }^2}{{\alpha }^2+{\beta }^2}}=\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}$

$z=\sqrt{{\alpha }^2+{\beta }^2}[\cos \varphi \sin \theta +\sin \varphi \cos \theta ]$

$z=\sqrt{{\alpha }^2+{\beta }^2}\sin (\varphi +\theta )$

$-1\le \sin (\varphi +\theta )\le 1$

$-\sqrt{{\alpha }^2+{\beta }^2}\le \sqrt{{\alpha }^2+{\beta }^2}\sin (\varphi +\theta )\le \sqrt{{\alpha }^2+{\beta }^2}$

$-\sqrt{{\alpha }^2+{\beta }^2}\le z\le \sqrt{{\alpha }^2+{\beta }^2}$

Therefore the maximum and minimum values of $z$ are $\sqrt{{\alpha }^2+{\beta }^2}$ and $\sqrt{{\alpha }^2+{\beta }^2}$ respectively.