Triple of sine

Geometry Level 3

( 3 − 4 sin ⁡ 2 9 ∘ ) ( 3 − 4 sin ⁡ 2 2 7 ∘ ) ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) = ? (3-4\sin^2{9^{\circ}})(3-4\sin^2{27^{\circ}})(3-4\sin^2{81^{\circ}})(3-4\sin^2{243^{\circ}}) = \ ?


The answer is 1.

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Ikkyu San by Ikkyu San , 23, Thailand

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1 solution

Nihar Mahajan
Nov 1, 2015

( 3 − 4 sin ⁡ 2 9 ) ( 3 − 4 sin ⁡ 2 27 ) ( 3 − 4 sin ⁡ 2 81 ) ( 3 − 4 sin ⁡ 2 243 ) = ( sin ⁡ 9 sin ⁡ 27 sin ⁡ 81 sin ⁡ 243 ) ( 3 − 4 sin ⁡ 2 9 ) ( 3 − 4 sin ⁡ 2 27 ) ( 3 − 4 sin ⁡ 2 81 ) ( 3 − 4 sin ⁡ 2 243 ) ( sin ⁡ 9 sin ⁡ 27 sin ⁡ 81 sin ⁡ 243 ) = ( 3 sin ⁡ 9 − 4 sin ⁡ 3 9 ) ( 3 sin ⁡ 27 − 4 sin ⁡ 3 27 ) ( 3 sin ⁡ 81 − 4 sin ⁡ 3 81 ) ( 3 sin ⁡ 243 − 4 sin ⁡ 3 243 ) ( sin ⁡ 9 sin ⁡ 27 sin ⁡ 81 sin ⁡ 243 ) = sin ⁡ 27 sin ⁡ 81 sin ⁡ 243 sin ⁡ 729 sin ⁡ 9 sin ⁡ 27 sin ⁡ 81 sin ⁡ 243 = sin ⁡ 729 sin ⁡ 9 = sin ⁡ 9 sin ⁡ 9 = 1 \large{\begin{aligned} & (3-4\sin^2{9})(3-4\sin^2{27})(3-4\sin^2{81})(3-4\sin^2{243}) \\ &= \dfrac{(\sin 9 \sin 27 \sin 81 \sin 243)(3-4\sin^2{9})(3-4\sin^2{27})(3-4\sin^2{81})(3-4\sin^2{243})}{(\sin 9 \sin 27 \sin 81 \sin 243)} \\&=\dfrac{(3\sin 9-4\sin^3{9})(3\sin 27-4\sin^3{27})(3\sin 81-4\sin^3{81})(3\sin 243-4\sin^3{243})}{(\sin 9 \sin 27 \sin 81 \sin 243)} \\ &= \dfrac{ \sin 27 \sin 81 \sin 243 \sin 729}{\sin 9 \sin 27 \sin 81 \sin 243}\\ &= \dfrac{\sin 729}{\sin 9} \\ &= \dfrac{\sin 9}{\sin 9} \\ &= \Large{\boxed{1}} \end{aligned}}

  • In third step I used the identity sin ⁡ ( 3 x ) = 3 sin ⁡ ( x ) − 4 sin ⁡ 3 ( x ) \sin(3x)=3\sin(x)-4\sin^3(x)
8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

When you see 3 − sin ⁡ 2 θ 3 - \sin^2 \theta , you should start thinking about the triple angle identity.

I accidentally hit the Discuss Solution button. \color{#D61F06}{\text{I accidentally hit the Discuss Solution button.}}

Let the product be P P , then we have:

P = ( 3 − 4 sin ⁡ 2 9 ∘ ) ( 3 − 4 sin ⁡ 2 2 7 ∘ ) ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) ⇒ sin ⁡ 9 ∘ P = sin ⁡ 9 ∘ ( 3 − 4 sin ⁡ 2 9 ∘ ) ( 3 − 4 sin ⁡ 2 2 7 ∘ ) ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) = ( 3 sin ⁡ 9 ∘ − 4 sin ⁡ 3 9 ∘ ) ( 3 − 4 sin ⁡ 2 2 7 ∘ ) ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) = sin ⁡ 2 7 ∘ ( 3 − 4 sin ⁡ 2 2 7 ∘ ) ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) sin ⁡ ( 3 θ ) = 3 sin ⁡ θ − 4 sin ⁡ 3 θ = sin ⁡ 8 1 ∘ ( 3 − 4 sin ⁡ 2 8 1 ∘ ) ( 3 − 4 sin ⁡ 2 24 3 ∘ ) = sin ⁡ 24 3 ∘ ( 3 − 4 sin ⁡ 2 24 3 ∘ ) = sin ⁡ 72 9 ∘ = sin ⁡ 9 ∘ ⇒ P = 1 \begin{aligned} P & = (3-4\sin^2 9^\circ)(3-4\sin^2 27^\circ) (3-4\sin^2 81^\circ) (3-4\sin^2 243^\circ) \\ \Rightarrow \sin 9^\circ P & = \sin 9^\circ (3-4\sin^2 9^\circ)(3-4\sin^2 27^\circ) (3-4\sin^2 81^\circ) (3-4\sin^2 243^\circ) \\ & = (\color{#3D99F6} {3\sin 9^\circ -4\sin^3 9^\circ})(3-4\sin^2 27^\circ) (3-4\sin^2 81^\circ) (3-4\sin^2 243^\circ) \\ & = \color{#3D99F6}{\sin 27^\circ} (3-4\sin^2 27^\circ) (3-4\sin^2 81^\circ) (3-4\sin^2 243^\circ) \quad \quad \small \color{#3D99F6}{\sin (3\theta) = 3 \sin \theta - 4\sin^3 \theta} \\ & = \sin 81^\circ (3-4\sin^2 81^\circ) (3-4\sin^2 243^\circ) \\ & = \sin 243^\circ (3-4\sin^2 243^\circ) \\ & = \sin 729^\circ \\ & = \sin 9^\circ \\ \Rightarrow P & = \boxed{1} \end{aligned}

Chew-Seong Cheong - 5 years, 7 months ago

sin 729 = sin 9 as period of sin is 360. 729 - 9 =720 = 2*360.

Noel Lo - 5 years, 7 months ago

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