A sum of coefficients.

Algebra Level 4

If sin ⁡ 2 A = x \sin ^{ 2 }{ A } = x , then

sin ⁡ A × sin ⁡ 2 A × sin ⁡ 3 A × sin ⁡ 4 A \sin { A } \times \sin { 2A } \times \sin { 3A } \times \sin { 4A }

is a polynomial in x x , the sum of whose coefficients is:

4 None -1 0

Raghav Vaidyanathan by Raghav Vaidyanathan , 23, USA

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2 solutions

Let f ( x ) = sin ⁡ A × sin ⁡ 2 A × sin ⁡ 3 A × sin ⁡ 4 A f(x)=\sin { A } \times \sin { 2A } \times \sin { 3A } \times \sin { 4A }

Then the required sum is simply the value of f ( 1 ) f(1) . Now, if x = 1 ⇒ sin ⁡ 2 A = 1 ⇒ sin ⁡ 2 A = 0 x=1\Rightarrow \sin ^{ 2 }{ A }=1 \Rightarrow \sin {2A}=0

⇒ f ( 1 ) = 0 \Rightarrow \boxed{ f(1)=0}

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Thanks for the solution.up voted. I am posting my long solution just as a poor alternative.

Niranjan Khanderia - 2 years, 10 months ago

S i n A = S i n A S i n 2 A = 2 S i n A ( C o s A ) S i n 3 A = S i n A ( 3 − 4 S i n 2 A ) = S i n A ( 3 − 4 x ) S i n 4 A = 4 S i n A ( C o s 3 A − S i n 2 ∗ C o s A ) = 4 S i n A ( C o s A ) ( 1 − x − x ) ∴ S i n A ∗ S i n 2 A ∗ S i n 3 A ∗ S i n 4 A = 8 ∗ S i n 4 x ( C o s 2 A ) ( 3 − 4 x ) ( 1 − 2 x ) = 8 ∗ x 2 ( 1 − x ) ( 3 − 4 x ) ( 1 − 2 x ) = 8 x 2 ( − 8 x 3 + 18 x 2 − 13 x + 3 ) − 8 + 18 − 13 + 3 = 0. \begin{aligned}\\ SinA&= SinA\\ Sin2A&= 2SinA(CosA)\\ Sin3A&= SinA(3-4Sin^2A)=SinA(3-4x)\\ Sin4A&= 4SinA(Cos^3A-Sin^2*CosA)=4SinA(CosA)(1-x-x)\\ ~~~ \\ ~~~~~\\ \therefore~~SinA*Sin2A*Sin3A*Sin4A&=8*Sin^4x(Cos^2A)(3-4x)(1-2x)\\ &=8*x^2(1-x)(3-4x)(1-2x)\\ &=8x^2(-8x^3+18x^2-13x+3)\\ \end{aligned} \\ ~~~ \\ ~~~~~\\ -8+18-13+3=\huge \color{#D61F06}{0}.

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