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# The value of $\dfrac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}}$ is equal toA. $cot3x$ B. $\cos 3x$ C. $\tan 3x$ D. None of these

Last updated date: 01st Mar 2024
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Hint: The value of the given trigonometric expression can be found by using the below formulas. Solve the numerator and the denominator separately and then put their results over each other to find the value. The ratio of cosine function over sine function is cotangent (cot) function and the inverse of cotangent is tangent function.
Formulas used:
$\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right),\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ , where x and y are the angles which can be equal or unequal.

We are given to find the value of a trigonometric expression $\dfrac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}}$
Let us consider the numerator first and it is
$\cos 4x + \cos 3x + \cos 2x$
Here as we can see there are three terms with cosine functions with different angle measures.
So let us consider the first and the third terms, $\cos 4x + \cos 2x$ .
On comparing the above two terms with $\cos x + \cos y$ , we get x is equal to 4x and y is equal to 2x and
$\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ .
On substituting x as 4x and y as 2x in the above formula, we get
$\cos 4x + \cos 2x = 2\cos \left( {\dfrac{{4x + 2x}}{2}} \right)\cos \left( {\dfrac{{4x - 2x}}{2}} \right)$
$\Rightarrow \cos 4x + \cos 2x = 2\cos \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{2x}}{2}} \right) = 2\cos 3x\cos x$
On substituting the value of $\cos 4x + \cos 2x$ in $\cos 4x + \cos 3x + \cos 2x$ , we get
$\cos 4x + \cos 3x + \cos 2x = \cos 4x + \cos 2x + \cos 3x = 2\cos 3x\cos x + \cos 3x$
$\therefore \cos 4x + \cos 3x + \cos 2x = \cos 3x\left( {2\cos x + 1} \right)$
Now, we are considering the numerator and it is $\sin 4x + \sin 3x + \sin 2x$
Here as we can see there are three terms with sine functions with different angle measures.
So let us consider the first and the third terms, $\sin 4x + \sin 2x$ .
On comparing the above two terms with $\sin x + \sin y$ , we get x is equal to 4x and y is equal to 2x and
$\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ .
On substituting x as 4x and y as 2x in the above formula, we get
$\sin 4x + \sin 2x = 2\sin \left( {\dfrac{{4x + 2x}}{2}} \right)\cos \left( {\dfrac{{4x - 2x}}{2}} \right)$
$\Rightarrow \sin 4x + \sin 2x = 2\sin \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{2x}}{2}} \right) = 2\sin 3x\cos x$
On substituting the value of $\sin 4x + \sin 2x$ in $\sin 4x + \sin 3x + \sin 2x$ , we get
$\Rightarrow \sin 4x + \sin 3x + \sin 2x = \sin 4x + \sin 2x + \sin 3x = 2\sin 3x\cos x + \sin 3x$
$\therefore \sin 4x + \sin 3x + \sin 2x = \sin 3x\left( {2\cos x + 1} \right)$
Now we are combining the values of numerator and denominator.
$\Rightarrow \dfrac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}} = \dfrac{{\cos 3x\left( {2\cos x + 1} \right)}}{{\sin 3x\left( {2\cos x + 1} \right)}} = \dfrac{{\cos 3x}}{{\sin 3x}}$
Here, we have got a ratio of cosine and sine functions, which is equal to the cotangent function (cot).
$\Rightarrow \dfrac{{\cos 3x}}{{\sin 3x}} = cot3x$
$\therefore \dfrac{{\cos 4x + \cos 3x + \cos 2x}}{{\sin 4x + \sin 3x + \sin 2x}} = cot3x$
Therefore, the correct option is Option A, $cot3x$ .
So, the correct answer is “Option A”.

Note: In the formula $\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ , we have both the right hand side terms as cosine functions whereas in the formula $\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ , both the right hand side terms are not sine functions (one is sine and other is cosine). So we should be careful while writing the formulas.