Q.The minimum value of $3\cos x + 4\sin x + 8$ is
(A) $5$
(B) $9$
(C) $7$
(D) $3$

Concept: Express $a\cos x + b\sin x$ as $R\cos(x - \phi)$ to find its range.

Step 1: Write $3\cos x + 4\sin x$ in the form $R\cos(x - \phi)$.

Here $R = \sqrt{3^2 + 4^2} = 5$.

Step 2: The range of $5\cos(x - \phi)$ is $[-5, 5]$.

So the minimum value of $3\cos x + 4\sin x$ is $-5$. …

The core idea here is that any expression of the form $a\cos x + b\sin x$ can be rewritten as a single sine (or cosine) function with an amplitude. This is a standard technique in trigonometry — it lets you find the maximum and minimum values directly without calculus.

Why does this work? Because $\cos x$ and $\sin x$ are orthogonal functions, and their linear combination traces out a circle in the $(a, b)$ plane. The maximum possible value of $a\cos x + b\sin x$ is $\sqrt{a^2 + b^2}$, and the minimum is $-\sqrt{a^2 + b^2}$. Adding a constant just shifts the entire range.

Let’s apply this to the given problem.

1. Identify the coefficients.

   We have $3\cos x + 4\sin x + 8$. Here $a = 3$ and $b = 4$, with a constant $c = 8$.

2. Find the amplitude $R$.

   The amplitude of $a\cos x + b\sin x$ is

3. Determine the range of the trigonometric part. Since $3\cos x + 4\sin x$ can be written as $R\sin(x + \phi)$ or $R\cos(x - \phi)$ for some phase $\phi$, its values lie between $-R$ and $R$. So:

4. Add the constant. Adding $8$ shifts the entire range upward:

which gives

5. Read off the minimum. …