Question

The maximum value of \(3\cos\theta + 4\sin\theta\) is

• A. -5
• B. 5
• C. 25
• D. None of these

Hint:

\(a\cos\theta + b\sin\theta\) lies between \(-\sqrt{a^2 + b^2}\) and \(\sqrt{a^2 + b^2}\).

Answer 1

The Correct Option is B

To determine the maximum value of the expression \(3\cos\theta + 4\sin\theta\), we can use the method of transforming it into the form \(R\cos(\theta + \alpha)\), where \(R\) is the amplitude of the expression.

First, assume the following identity:

\(R\cos(\theta + \alpha) = R(\cos\theta \cos\alpha - \sin\theta \sin\alpha)\)

Comparing this with \(3\cos\theta + 4\sin\theta\), we have:

• \(\cos\alpha = \frac{3}{R}\)
• \(\sin\alpha = \frac{4}{R}\)

Using the identity \((\cos\alpha)^2 + (\sin\alpha)^2 = 1\), we can find \(R\):

\((\frac{3}{R})^2 + (\frac{4}{R})^2 = 1\)

Solving this, we get:

\(\frac{9}{R^2} + \frac{16}{R^2} = 1\)

Combine the fractions:

\(\frac{25}{R^2} = 1\)

This implies:

\(R^2 = 25 \quad \Rightarrow \quad R = 5\)

Therefore, the maximum value of the expression \(3\cos\theta + 4\sin\theta\) is given by the amplitude \(R\), which is 5.

Thus, the correct answer is:

• Option: 5