Question

If ${\sin ^3}x \cdot \sin 3x = \sum\limits_{m = 0}^n {{C_m}} \cos mx$ where ${C_0},{C_1},{C_2},......{C_n}$ are constants and ${C_n} \ne 0$ then $n = $

Answer 1

Hint: Here, we will first of all, solve the LHS and express it in the form of summation of different angles of cosine function. Since, in the RHS, We have cosine function which can have angles up to $nx$, we will compare it with the LHS. The highest angle present in the LHS will be the required value of the highest possible angle of cosine function, i.e. $nx$. Comparing both the angles, we will find the value of $n$.

Formula Used:
 We will use the following formulas:
1.\[\sin 3x = 3\sin x - 4{\sin ^3}x\]
2.$\cos \left( {A - B} \right) - \cos \left( {A + B} \right) = 2\sin A\sin B$

Complete step-by-step answer:
It is given that ${\sin ^3}x \cdot \sin 3x = \sum\limits_{m = 0}^n {{C_m}} \cos mx$.
Now, first of all, we will solve the LHS.
LHS$ = {\sin ^3}x \cdot \sin 3x$
Here, using the formula \[\sin 3x = 3\sin x - 4{\sin ^3}x\], we get
\[ \Rightarrow 4{\sin ^3}x = 3\sin x - \sin 3x\]
Dividing both sides by 4, we get
\[ \Rightarrow {\sin ^3}x = \dfrac{1}{4}\left( {3\sin x - \sin 3x} \right)\]
Hence, substituting this value in the LHS, we get,
${\sin ^3}x \cdot \sin 3x = \dfrac{1}{4}\left( {3\sin x - \sin 3x} \right)\sin 3x$
Now, opening the bracket,
$ \Rightarrow {\sin ^3}x \cdot \sin 3x = \dfrac{1}{4}\left( {3\sin x \cdot \sin 3x - \sin 3x \cdot \sin 3x} \right)$
Multiplying and dividing by 2,
$ \Rightarrow {\sin ^3}x \cdot \sin 3x = \dfrac{1}{8}\left[ {3 \times \left( {2\sin 3x \cdot \sin x} \right) - 2\sin 3x \cdot \sin 3x} \right]$
Now, using the formula:
$\cos \left( {A - B} \right) - \cos \left( {A + B} \right) = 2\sin A\sin B$
We get,
$ \Rightarrow {\sin ^3}x \cdot \sin 3x = \dfrac{1}{8}\left[ {3 \times \left( {\cos \left( {3x - x} \right) - \cos \left( {3x + x} \right)} \right) - \left( {\cos \left( {3x - 3x} \right) - \cos \left( {3x + 3x} \right)} \right)} \right]$
$ \Rightarrow {\sin ^3}x \cdot \sin 3x = \dfrac{1}{8}\left[ {3 \times \left( {\cos 2x - \cos 4x} \right) - \left( {\cos 0x - \cos 6x} \right)} \right]$
Now, opening the brackets,
$ \Rightarrow {\sin ^3}x \cdot \sin 3x = \dfrac{1}{8}\left[ {3\cos 2x - 3\cos 4x - \cos 0x + \cos 6x} \right]$……………………………….$\left( 1 \right)$
Now, RHS$ = \sum\limits_{m = 0}^n {{C_m}} \cos mx$
Putting $m = 0$, we get,
\[{C_0}\cos 0x\]
Putting $m = 1$, we get,
\[{C_1}\cos x\]
Similarly, we can get the other values.
Thus we can have,
\[{C_0}\cos 0x,{C_1}\cos x,{C_2}\cos 2x,.....{C_6}\cos 6x\]
Where, ${C_0},{C_1},{C_2},......{C_n}$ are constants and ${C_n} \ne 0$
Hence, on comparing this with $\left( 1 \right)$, we can see that the highest possible angle, i.e. \[nx = 6x\]
Thus, the value of $n = 6$
Therefore, if ${\sin ^3}x \cdot \sin 3x = \sum\limits_{m = 0}^n {{C_m}} \cos mx$ where ${C_0},{C_1},{C_2},......{C_n}$ are constants and ${C_n} \ne 0$ then $n = 6$
Thus, this is the required answer.

Note: This question involved Trigonometry which is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In the simple terms they are written as ‘sin’, ‘cos’ and ‘tan’.