Question

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

Answer 1

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.