Question
What is the maximum value of $a \sin x+b \cos x$ ?

Answer

Suppose
$
\begin{aligned}
y & =a \sin x+b \sin x \\
\frac{d y}{d x} & =a \cos x-b \sin x
\end{aligned}
$
For, maxima and minima $\frac{d y}{d x}=0$$
\begin{array}{l}
\therefore a \cos x-b \sin x=0 \\
\tan x=\frac{a}{b} \\
\frac{d^2 y}{d x^2}=-a \sin x-b \cos x \\
=-(a \sin x+b \cos x)=-ve \\
\therefore \text { maximum value }=\frac{a \times a}{\sqrt{a^2+b^2}}+\frac{b \times b}{\sqrt{a^2+b^2}} \\
=\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}
\end{array}
$

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