Trignometry - 6

Geometry Level 4

sin ⁡ ( A ) sin ⁡ ( 2 A ) sin ⁡ ( 3 A ) sin ⁡ ( 4 A ) = a x 2 + b x 3 + c x 4 + d x 5 \sin (A) \sin(2A) \sin(3A) \sin(4A) = ax^2 + bx^3 + cx^4 + dx^5

If sin ⁡ 2 ( A ) = x \sin^{2}(A)= x and that the equation above is satisfied for constants a , b , c , d a,b,c,d , then find the value of 12 a − 10 b + 8 c − 11 d 12a - 10b +8c -11d .


The answer is 3184.

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Kunal Joshi by Kunal Joshi , 24, India

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1 solution

Chew-Seong Cheong
Oct 15, 2014

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

= sin ⁡ A ( 2 sin ⁡ A cos ⁡ A ) ( sin ⁡ A cos ⁡ 2 A + sin ⁡ 2 A cos ⁡ A ) ( 2 sin ⁡ 2 A cos ⁡ 2 A ) =\sin{A} (2\sin{A}\cos{A})(\sin{A}\cos{2A}+\sin{2A}\cos{A})(2\sin{2A}\cos{2A})

= sin ⁡ A ( 2 sin ⁡ A cos ⁡ A ) [ sin ⁡ A ( 1 − 2 sin ⁡ 2 A ) + 2 sin ⁡ A cos ⁡ 2 A ) ] [ 4 sin ⁡ A cos ⁡ A ( 1 − 2 sin ⁡ 2 A ) ] =\sin{A} (2\sin{A}\cos{A})[\sin{A}(1-2\sin^2{A})+2\sin{A}\cos^2{A})][4\sin{A}\cos{A}(1-2\sin^2{A})]

= 8 sin ⁡ 4 A cos ⁡ 2 A ( 1 − 2 sin ⁡ 2 A + 2 cos ⁡ 2 A ) ( 1 − 2 sin ⁡ 2 A ) =8\sin^4{A} \cos^2{A}(1-2\sin^2{A}+2\cos^2{A})(1-2\sin^2{A})

= 8 x 2 ( 1 − x ) ( 1 − 2 x + 2 − 2 x ) ( 1 − 2 x ) = 8 x 2 ( 1 − x ) ( 3 − 4 x ) ( 1 − 2 x ) =8x^2 (1-x) (1-2x+2 - 2x )(1-2x) = 8x^2 (1-x) (3-4x)(1-2x)

= 8 x 2 ( 3 − 7 x + 4 x 2 ) ( 1 − 2 x ) = 8 x 2 ( 3 − 7 x + 4 x 2 − 6 x + 14 x − 8 x 2 ) =8x^2 (3 -7x +4x^2)(1-2x) = 8x^2 (3 -7x +4x^2 - 6x +14x -8x^2)

= 8 x 2 ( 3 − 13 x + 18 x 2 − 8 x 3 ) = 24 x ∗ 2 − 104 x 3 + 144 x 4 − 64 x 5 = 8x^2 (3 -13x +18x^2 -8x^3) = 24x*2 -104x^3+144x^4-64x^5

⇒ a = 24. b = − 104 , c = 144 , d = − 64 \Rightarrow a = 24. b= -104, c =144, d =-64

⇒ 12 a − 10 b + 8 c − 11 d = 288 + 1040 + 1152 + 704 = 3184 \Rightarrow 12a -10b + 8c -11 d = 288+1040+1152 +704=\boxed{3184}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, could you please explain how sin ⁡ 3 A = sin ⁡ A cos ⁡ 2 A + sin ⁡ 2 A cos ⁡ A \sin { 3A } =\sin { A } \cos { 2A } +\sin { 2A } \cos { A } came through?

Anandhu Raj - 6 years ago

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sin ⁡ ( A + B ) = sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A \sin{(A+B)} = \sin{A}\cos{B}+\sin{B}\cos{A} replace B B with 2 A 2A .

Chew-Seong Cheong - 6 years ago

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Ew! I was so dumb that I failed to notice that :(

Anandhu Raj - 6 years ago

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@Anandhu Raj – It is okay. Sometimes we are like that.

Chew-Seong Cheong - 6 years ago

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