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Question: If \( {\sin ^2}A = x \) , then \( \sin A\sin 2A\sin 3A\sin 4A \) is a polynomial in \( x \) , the su...

If sin⁡2A=x{\sin ^2}A = x , then sin⁡Asin⁡2Asin⁡3Asin⁡4A\sin A\sin 2A\sin 3A\sin 4A is a polynomial in xx , the sum of whose coefficient is
a. 0
b. 40
c. 168
d. 336

Explanation

Solution

Hint : Here in this question first we multiply the sin⁡Asin⁡2Asin⁡3Asin⁡4A\sin A\sin 2A\sin 3A\sin 4A and then we are going to substitute the needed values to the equation and we are adding the coefficient of the equation and hence we obtain the required result.

Complete step-by-step answer :
Now consider the polynomial sin⁡Asin⁡2Asin⁡3Asin⁡4A\sin A\sin 2A\sin 3A\sin 4A
Where sin⁡2A=2sin⁡A.cos⁡A\sin 2A = 2\sin A.\cos A , sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4{\sin ^3}A and
sin⁡4A=sin⁡(3A+A)\sin 4A = \sin (3A + A)
Now we apply the formula sin⁡(A+B)=sin⁡A.cos⁡B+cos⁡A.sin⁡B\sin (A + B) = \sin A.\cos B + \cos A.\sin B we have
⇒sin⁡4A=sin⁡3A.cos⁡A+cos⁡3A.sin⁡A\Rightarrow \sin 4A = \sin 3A.\cos A + \cos 3A.\sin A
And by applying the formula sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4{\sin ^3}A and cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4{\cos ^3}A - 3\cos A , hence we have ⇒sin⁡4A=(3sin⁡A−4sin⁡3A).cos⁡A+(4cos⁡3A−3cos⁡A).sin⁡A\Rightarrow \sin 4A = (3\sin A - 4{\sin ^3}A).\cos A + (4{\cos ^3}A - 3\cos A).\sin A
⇒sin⁡4A=(3sin⁡Acos⁡A−4sin⁡3Acos⁡A)+(4cos⁡3Asin⁡A−3cos⁡Asin⁡A)\Rightarrow \sin 4A = (3\sin A\cos A - 4{\sin ^3}A\cos A) + (4{\cos ^3}A\sin A - 3\cos A\sin A)
On simplification
⇒sin⁡4A=4cos⁡3Asin⁡A−4sin⁡3Acos⁡A\Rightarrow \sin 4A = 4{\cos ^3}A\sin A - 4{\sin ^3}A\cos A
Now we will substitute all the values in the polynomial we have
sin⁡Asin⁡2Asin⁡3Asin⁡4A=sin⁡A(2sin⁡Acos⁡A)(3sin⁡A−4sin⁡3A)(4cos⁡3Asin⁡A−4sin⁡3Acos⁡A)\sin A\sin 2A\sin 3A\sin 4A = \sin A(2\sin A\cos A)(3\sin A - 4{\sin ^3}A)(4{\cos ^3}A\sin A - 4{\sin ^3}A\cos A)
On multiplying the first two terms we have
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=(2sin⁡2Acos⁡A)(3sin⁡A−4sin⁡3A)(4cos⁡3Asin⁡A−4sin⁡3Acos⁡A)\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = (2{\sin ^2}A\cos A)(3\sin A - 4{\sin ^3}A)(4{\cos ^3}A\sin A - 4{\sin ^3}A\cos A)
Multiply the first two terms of the above equation
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=(6sin⁡3Acos⁡A−8sin⁡5Acos⁡A)(4cos⁡3Asin⁡A−4sin⁡3Acos⁡A)\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = (6{\sin ^3}A\cos A - 8{\sin ^5}A\cos A)(4{\cos ^3}A\sin A - 4{\sin ^3}A\cos A)
On further simplification we have
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24sin⁡4Acos⁡4A−24sin⁡6Acos⁡2A−32sin⁡6Acos⁡4A+32sin⁡8Acos⁡2A\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{\sin ^4}A{\cos ^4}A - 24{\sin ^6}A{\cos ^2}A - 32{\sin ^6}A{\cos ^4}A + 32{\sin ^8}A{\cos ^2}A
By the data we have sin⁡2A=x{\sin ^2}A = x applying the square root we have
sin⁡A=±x\sin A = \pm x
We know the standard trigonometric equation that is sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 , this can be written as cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x . As we know the value of the sin⁡2A=x{\sin ^2}A = x and using the standard trigonometric equations we can write as cos⁡2A=1−x{\cos ^2}A = 1 - x .
Now consider the polynomial that is sin⁡Asin⁡2Asin⁡3Asin⁡4A=24sin⁡4Acos⁡4A−24sin⁡6Acos⁡2A−32sin⁡6Acos⁡4A+32sin⁡8Acos⁡2A\sin A\sin 2A\sin 3A\sin 4A = 24{\sin ^4}A{\cos ^4}A - 24{\sin ^6}A{\cos ^2}A - 32{\sin ^6}A{\cos ^4}A + 32{\sin ^8}A{\cos ^2}A
Substituting the values, we have
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24(x)4(1−x)2−24(x)6(1−x)−32(x)6(1−x)2+32(x)8(1−x)\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{(x)^4}{(1 - x)^2} - 24{(x)^6}(1 - x) - 32{(x)^6}{(1 - x)^2} + 32{(x)^8}(1 - x)
Here we use (a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2} to the (1−x)2{(1 - x)^2}
Hence, we have
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24x4(1−2x+x2)−24x6(1−x)−32x6(1−2x+x2)+32x8(1−x)\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{x^4}(1 - 2x + {x^2}) - 24{x^6}(1 - x) - 32{x^6}(1 - 2x + {x^2}) + 32{x^8}(1 - x)
On further simplification
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24x4−48x5+24x6−24x6+24x7−32x6+64x7−32x8+32x8−32x9\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{x^4} - 48{x^5} + 24{x^6} - 24{x^6} + 24{x^7} - 32{x^6} + 64{x^7} - 32{x^8} + 32{x^8} - 32{x^9}
Cancelling some terms
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24x4−48x5+24x7−32x6+64x7−32x9\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{x^4} - 48{x^5} + 24{x^7} - 32{x^6} + 64{x^7} - 32{x^9}
Adding the terms which are having same power raised to x,
⇒sin⁡Asin⁡2Asin⁡3Asin⁡4A=24x4−48x5+88x7−32x6−32x9\Rightarrow \sin A\sin 2A\sin 3A\sin 4A = 24{x^4} - 48{x^5} + 88{x^7} - 32{x^6} - 32{x^9}
Now we have to the only coefficient of the polynomial, here we have the 24, -48, 88, -32 and -32
Now we add the coefficients 24−48+88−32−3224 - 48 + 88 - 32 - 32
We will add 24 and -48
⇒−24+88−32−32\Rightarrow - 24 + 88 - 32 - 32
Now we will add -24 and 88
⇒64−32−32\Rightarrow 64 - 32 - 32
Now we will add 64 and -32
⇒32−32=0\Rightarrow 32 - 32 = 0 and hence we have 0 has the sum of coefficients of the polynomial.
So, the correct answer is “Option a”.

Note : We must know about the trigonometry formulas and conditions and we should multiply the term very carefully. Since the polynomial contains four terms if we go wrong in any one of the steps, we may not be able to obtain the correct solution for the question.