Tricky Trigonometry - 3

Geometry Level 4

If sin ⁡ 2 θ = x \sin^2\theta = x and ∏ r = 1 4 sin ⁡ ( r θ ) = a x 2 + b x 3 + c x 4 + d x 5 \displaystyle \prod^{4}_{r = 1} \sin(r\theta) = ax^2 + bx^3 + cx^4 + dx^5 , then find the value of a + b + c + d a + b + c + d .

Note: ∏ r = 1 n r = 1 ⋅ 2 ⋅ 3 . . . . n \displaystyle \prod^{n}_{r = 1} r = 1 \cdot 2 \cdot 3 \quad .... \quad n

Also try Tricky Trigonometry-1 and Tricky Trigonometry-2


The answer is 0.

Ram Mohith by Ram Mohith , 18, India

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3 solutions

∏ r = 1 4 sin ⁡ ( r θ ) = sin ⁡ θ sin ⁡ 2 θ sin ⁡ 3 θ sin ⁡ 4 θ = sin ⁡ θ ( 2 sin ⁡ θ cos ⁡ θ ) ( 3 sin ⁡ θ − 4 sin ⁡ 3 θ ) ( 2 ( 2 sin ⁡ θ cos ⁡ θ ) ( 1 − 2 sin ⁡ 2 θ ) ) = 8 sin ⁡ 2 θ cos ⁡ 2 θ ( 3 sin ⁡ 2 θ − 4 sin ⁡ 4 θ ) ( 1 − 2 sin ⁡ 2 θ ) Given that sin ⁡ 2 θ = x = 8 x ( 1 − x ) ( 3 x − 4 x 2 ) ( 1 − 2 x ) ⟹ cos ⁡ 2 θ = 1 − x = 24 x 2 − 104 x 3 + 144 x 4 − 64 x 5 \begin{aligned} \prod_{r=1}^4 \sin (r\theta) & = \sin \theta \sin 2\theta \sin 3\theta \sin 4 \theta \\ & = \sin \theta (2\sin \theta \cos \theta) (3\sin \theta - 4\sin^3 \theta )\left(2(2\sin \theta \cos \theta)(1-2\sin^2\theta)\right) \\ & = 8 \sin^2 \theta \cos^2 \theta (3\sin^2 \theta - 4\sin^4 \theta )(1-2\sin^2\theta) & \small \blue{\text{Given that }\sin^2 \theta = x} \\ & = 8 x(1-x)(3x - 4x^2)(1-2x) & \small \blue{\implies \cos^2 \theta = 1-x} \\ & = 24x^2 -104x^3 + 144x^4-64x^5 \end{aligned}

Therefore, a + b + c + d = 24 − 104 + 144 − 64 = 0 a+b+c+d=24-104+144-64 = \boxed 0 .

3 Helpful 2 Interesting 2 Brilliant 0 Confused
Jon Haussmann
Oct 10, 2019

Let θ = π 2 \theta = \frac{\pi}{2} . Then sin ⁡ θ sin ⁡ 2 θ sin ⁡ 3 θ sin ⁡ 4 θ = 0. \sin \theta \sin 2 \theta \sin 3 \theta \sin 4 \theta = 0. Also, x = 1 x = 1 , so a x 2 + b x 3 + c x 4 + d x 5 = a + b + c + d ax^2 + bx^3 + cx^4 + dx^5 = a + b + c + d . Therefore, a + b + c + d = 0 a + b + c + d = 0 .

0 Helpful 1 Interesting 2 Brilliant 0 Confused

A good and faster method for objective type questions.

Ram Mohith - 1 year, 8 months ago

From the given data of the problem we get sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ = 24 x 2 − 104 x 3 + 144 x 4 − 64 x 5 \sin \theta \times\sin 2\theta \times\sin 3\theta \times\sin 4\theta =24x^2-104x^3+144x^4-64x^5 . So a = 24 , b = − 104 , c = 144 , d = − 64 a=24,b=-104,c=144,d=-64 and a + b + c + d = 0 a+b+c+d=\boxed{0}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

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