The maximum value of the function \[3\cos x - 4\sin x\] is
A. \[2\]
B. \[3\]
C. \[4\]
D. \[5\]

Hint: In this question we have to find the maximum value of the given function. We will first simplify the given function \[3\cos x - 4\sin x\] in terms of sine function by multiplying and dividing simultaneously by \[5\] . Then we will use the fact that the maximum value of sine function is \[1\] to get the desired result.
Formula used: \[\sin (y - x) = \sin y\cos x - \cos y\sin x\]

Complete answer:
This problem is based on trigonometric identities. Trigonometry is the branch of mathematics that deals with triangles, its sides, and angles . While trigonometric identities are those combinations of constants and t-ratios of angles that is true for different values of angles. For example, \[{\sin ^2}x + {\cos ^2}x = 1\] is a trigonometric identity.
Consider the given question,
The given function is \[3\cos x - 4\sin x\].
Let \[f(x) = 3\cos x - 4\sin x\].
Multiplying and dividing the above function by \[\sqrt {{3^2} + {{( - 4)}^2}} \](i.e. \[5\]) we have ,
\[
  f(x) = \dfrac{5}{5}\left( {3\cos x - 4\sin x} \right) \\
  f(x) = 5\left( {\dfrac{3}{5}\cos x - \dfrac{4}{5}\sin x} \right) \\
\]
Now consider, \[\sin y = \dfrac{3}{5}\] , then \[\cos y = \dfrac{4}{5}\].
Hence from above we have,
\[f(x) = 5\left( {\sin y\cos x - \cos y\sin x} \right)\]
We know that, \[\sin (y - x) = \sin y\cos x - \cos y\sin x\].
Hence we have,
\[f(x) = 5\sin (y - x)\]
Thus we have \[f(x) = 3\cos x - 4\sin x = 5\sin (y - x)\]
Now we have to find the maximum value of the function.
We know that the value of sine function lies between \[ - 1\] and \[1\].
i.e. \[ - 1 \leqslant \sin (y - x) \leqslant 1\]
Multiplying the above inequality by 5 we get,
i.e. \[ - 5 \leqslant 5\sin (y - x) \leqslant 5\]
i.e. \[ - 5 \leqslant f(x) \leqslant 5\]
Therefore, the maximum value of the function \[3\cos x - 4\sin x\] is \[5\].
Hence Option D is correct.

Note:
To solve the trigonometric identity of the form \[a\cos x + b\sin x\], we multiply and divide by \[\sqrt {{a^2} + {b^2}} \] where \[a\] is the coefficient of \[\cos x\] and \[b\] is the coefficient of \[\sin x\].
i.e. \[a\cos x + b\sin x = \sqrt {{a^2} + {b^2}} \left( {\dfrac{a}{{\sqrt {{a^2} + {b^2}} }}\cos x + \dfrac{b}{{\sqrt {{a^2} + {b^2}} }}\sin x} \right)\]
Now we consider \[\dfrac{a}{{\sqrt {{a^2} + {b^2}} }} = \sin y\]
This implies \[\dfrac{b}{{\sqrt {{a^2} + {b^2}} }} = \cos y\]
Hence we have,
\[a\cos x + b\sin x = \sqrt {{a^2} + {b^2}} \left( {\sin y\cos x + \cos y\sin x} \right)\]
On solving, we have,
\[a\cos x + b\sin x = \sqrt {{a^2} + {b^2}} \sin (x + y)\].