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# Find the range of $13\cos x + 3\sqrt 3 \sin x - 4$

Last updated date: 19th Jun 2024
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Hint: Observe that${13^2} + {(3\sqrt 3 )^2} = {14^2}$.
Multiply and divide the given function by 14 to get a point on the unit circle. Also, use the fact: For every point $P(x,y)$on the unit circle${x^2} + {y^2} = 1$there exists $\theta \in [0,2\pi )$such that$x = \cos \theta$ and $y = \sin \theta$.
Simplify the function to get an expression without $\sin x$using the trigonometric identity $\cos (A - B) = \cos A\cos B - \sin A\sin B$. Finally, use the range of $\cos \theta$ : $- 1 \leqslant \cos (x - \theta ) \leqslant 1$ and get the answer.

Complete step by step solution:
We are given a trigonometric function $13\cos x + 3\sqrt 3 \sin x - 4$
We need to determine the range of this function.
Let $f(x) = 13\cos x + 3\sqrt 3 \sin x - 4$
A range of a function $f$would be the set of all the outcomes or outputs for the various inputs in its domain. It is
Domain of a function $f$is the set of all possible values on which $f$can be applied.
This means if $x$is a variable, then it is possible that for some values of $x$, the function is not defined.
We would be simplifying the given function to ease the process of finding the range.
We can see that${13^2} + {(3\sqrt 3 )^2} = 169 + 27 = 196 = {14^2}$
Therefore, we multiply and divide by 14 throughout the expression of the given function
Then, we get
$f(x) = 13\cos x + 3\sqrt 3 \sin x - 4 \\ = 14(\dfrac{{13}}{{14}}\cos x + \dfrac{{3\sqrt 3 }}{{14}}\sin x) - 4..............(1) \\$
Now, we can observe that ${(\dfrac{{13}}{{14}})^2} + {(\dfrac{{3\sqrt 3 }}{{14}})^2} = \dfrac{{169}}{{196}} + \dfrac{{27}}{{196}} = \dfrac{{196}}{{196}} = 1$
This implies that $(\dfrac{{13}}{{14}},\dfrac{{3\sqrt 3 }}{{14}})$is a point on the unit circle${x^2} + {y^2} = 1$
Let us recall a fact here: For every point $P(x,y)$on the unit circle${x^2} + {y^2} = 1$there exists $\theta \in [0,2\pi )$ such that $x = \cos \theta$ and $y = \sin \theta$.
Therefore, for the point $(\dfrac{{13}}{{14}},\dfrac{{3\sqrt 3 }}{{14}})$, there exists $\theta \in [0,2\pi )$such that $\dfrac{{13}}{{14}} = \cos \theta$ and $\dfrac{{3\sqrt 3 }}{{14}} = \sin \theta$.
Then, on substituting in equation (1), we get
$f(x) = 14(\cos \theta \cos x + \sin \theta \sin x) - 4 \\ = 14(\cos x\cos \theta + \sin x\sin \theta ) - 4 \\$
Here we have rearranged the cosine and sine values. We can do this because they are real numbers.
We will use the angle difference identity:
$\cos (A - B) = \cos A\cos B - \sin A\sin B$
Then$f(x) = 14(\cos (x - \theta )) - 4$
Now, we know that the range of $\cos \theta$ is $[ - 1,1]$ for any angle $\theta$
That is, $- 1 \leqslant \cos \theta \leqslant 1$ for any angle $\theta$
Therefore,
$- 1 \leqslant \cos (x - \theta ) \leqslant 1$
Multiplying 14 throughout the expression, we get
$- 14 \leqslant 14\cos (x - \theta ) \leqslant 14$
Subtracting 4 from each value, we get
$- 14 + 4 \leqslant 14\cos (x - \theta ) + 4 \leqslant 14 + 4 \\ \Rightarrow - 18 \leqslant 14\cos (x - \theta ) + 4 \leqslant 18 \\$
Hence, the range of $f(x) = 13\cos x + 3\sqrt 3 \sin x - 4$ is $[ - 18,18]$.

Note: The inequality in the final step means that every real number between -18 and 18 belongs to the range of $f(x) = 13\cos x + 3\sqrt 3 \sin x - 4$. Therefore, writing {-18, 18} as an answer is completely wrong.
Here, $[ - 18,18]$ indicates the closed interval taking every value from -18 to 18.