Question

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

Answer 1

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,
$a\cos x + b\sin x = \sqrt {{a^2} + {b^2}} \sin (x + y)$.