The maximum value of f(x)=5cos 3x+12sin 3x is:

The correct answer is Option (2) → 13

Here \(\sqrt{5^2+12^2}=\sqrt{169}=13\). Taking 13 out:

    \( \displaystyle f(x) = 13\left(\frac{5}{13}\cos 3x+\frac{12}{13}\sin 3x\right) \)

Let \(\cos\alpha=\frac{5}{13}\), \(\sin\alpha=\frac{12}{13}\) (possible since \(\left(\frac{5}{13}\right)^2+\left(\frac{12}{13}\right)^2=1\)):

    \(f(x) = 13(\cos\alpha\cos 3x+\sin\alpha\sin 3x)\)
    \(f(x) = 13\cos(3x-\alpha)\)

Since the largest value of the cosine is 1, reached when \(3x=\alpha\), the maximum value is 13.

Tip: The maximum of \(a\cos\theta+b\sin\theta\) is \(\sqrt{a^2+b^2}\) and the minimum is \(-\sqrt{a^2+b^2}\), whatever multiple of x the angle is.