Answer
418.2k+ views
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.
Complete step-by-step answer:
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.
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.
Complete step-by-step answer:
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)