Solveeit Logo
Login

Question

Question: If \({{\sin }^{3}}x\sin 3x=\sum\limits_{m=0}^{n}{{{C}_{m}}\cos mx}\) is an identity in \(x\), where ...

If sin⁡3xsin⁡3x=∑m=0nCmcos⁡mx{{\sin }^{3}}x\sin 3x=\sum\limits_{m=0}^{n}{{{C}_{m}}\cos mx} is an identity in xx, where C0,C1...Cn{{C}_{0}},{{C}_{1}}...{{C}_{n}} are constants and Cn≠0{{C}_{n}}\ne 0, then the value of nn is
(1) 2\left( 1 \right)\text{ }2
(2) 4\left( 2 \right)\text{ 4}
(3) 6\left( 3 \right)\text{ 6}
(4) 8\left( 4 \right)\text{ 8}

Explanation

Solution

In this question we have been given with a trigonometric expression on the left-hand side, which is equated with a series of the right-hand side. We will solve this question by first writing the left-hand side and expanding it using the trigonometric formula sin⁡3x=3sin⁡x−sin⁡3x4{{\sin }^{3}}x=\dfrac{3\sin x-\sin 3x}{4} and substitute it in the expression. We will also use the angle addition subtraction formula to convert the equation in terms of cos⁡x\cos x and then use comparison to find the value of nn.

Complete step by step answer:
We have the expression given to us as:
⇒sin⁡3xsin⁡3x=∑m=0nCmcos⁡mx\Rightarrow {{\sin }^{3}}x\sin 3x=\sum\limits_{m=0}^{n}{{{C}_{m}}\cos mx}
Consider the left-hand side of the expression, we get:
⇒sin⁡3xsin⁡3x\Rightarrow {{\sin }^{3}}x\sin 3x
Now we know the formula that sin⁡3x=3sin⁡x−sin⁡3x4{{\sin }^{3}}x=\dfrac{3\sin x-\sin 3x}{4} therefore, on substituting, we get:
⇒(3sin⁡x−sin⁡3x4)sin⁡3x\Rightarrow \left( \dfrac{3\sin x-\sin 3x}{4} \right)\sin 3x
On multiplying the terms, we get:
⇒14(3sin⁡xsin⁡3x−sin⁡3xsin⁡3x)\Rightarrow \dfrac{1}{4}\left( 3\sin x\sin 3x-\sin 3x\sin 3x \right)
Now we can see that the expression is in the form of sin⁡x\sin x. We will convert the expression in the form of cos⁡x\cos x.
We know that cos⁡(A+B)−cos⁡(A−B)=−2sin⁡Asin⁡B\cos \left( A+B \right)-\cos \left( A-B \right)=-2\sin A\sin B
Therefore, we can write sin⁡Asin⁡B=−12[cos⁡(A+B)−cos⁡(A−B)]\sin A\sin B=-\dfrac{1}{2}\left[ \cos \left( A+B \right)-\cos \left( A-B \right) \right]
On substituting in the expression and taking the terms common, we get:
⇒14(−32(cos⁡(4x)−cos⁡(−2x))+12(cos⁡(6x)−cos⁡(0x)))\Rightarrow \dfrac{1}{4}\left( -\dfrac{3}{2}\left( \cos \left( 4x \right)-\cos \left( -2x \right) \right)+\dfrac{1}{2}\left( \cos \left( 6x \right)-\cos \left( 0x \right) \right) \right)
On taking 12\dfrac{1}{2} common and simplifying, we get:
⇒18(cos⁡6x−cos⁡0x+3cos⁡2x−3cos⁡4x)\Rightarrow \dfrac{1}{8}\left( \cos 6x-\cos 0x+3\cos 2x-3\cos 4x \right), which is the required expansion for the left-hand side.
Now consider the right-hand side, we get:
⇒∑m=0nCmcos⁡mx\Rightarrow \sum\limits_{m=0}^{n}{{{C}_{m}}\cos mx}
On supposing the terms from 1,2,31,2,3 onwards, we get the series as:
⇒C1cos⁡x+C2cos⁡2x+C3cos⁡3x\Rightarrow {{C}_{1}}\cos x+{{C}_{2}}\cos 2x+{{C}_{3}}\cos 3x.
Now on comparing the series with the left-hand side we can see that the greatest value of mm present is 66 therefore, we can write:
⇒n=6\Rightarrow n=6, which Is the required solution.

So, the correct answer is “Option 3”.

Note: The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions. To simplify any given equation, it is good practice to convert all the identities into sin⁡x\sin x and cos⁡x\cos x for simplifying. If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.