A Sine Expression

Geometry Level 4

The function sin ( 5 θ ) \sin (5\theta) can be written of the form a sin 5 θ b sin 3 θ + c sin θ a\sin^5 \theta - b \sin^3 \theta +c \sin \theta

Find the value of a + b + c 14 a+b+c-14

This problem is inspired by this one and belongs to this set .


The answer is 27.

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1 solution

Chew-Seong Cheong
Jul 18, 2015

Using the Euler's identity:

e i 5 θ = ( e i θ ) 5 cos 5 θ + i sin 5 θ = ( cos θ + i sin θ ) 5 sin 5 θ = ( cos θ + i sin θ ) 5 [ Let a = cos θ and b = sin θ ] = ( a + i b ) 5 = ( a 5 + i 5 a 4 b 10 a 3 b 2 i 10 a 2 b 3 + 5 a b 4 + i b 5 ) = 5 a 4 b 10 a 2 b 3 + b 5 = 5 ( 1 b 2 ) 2 b 10 ( 1 b 2 ) b 3 + b 5 = 5 ( 1 2 b 2 + b 4 ) b 10 b 3 + 10 b 5 + b 5 = 5 b 10 b 3 + 5 b 5 10 b 3 + 10 b 5 + b 5 = 16 b 5 20 b 3 + 5 b = 16 sin 5 θ 20 sin 3 θ + 5 sin θ \begin{aligned} e^{i5\theta} & = \left(e^{i\theta} \right)^5 \\ \cos{5\theta} + i \sin{5\theta} & = (\cos{\theta} + i \sin{\theta})^5 \\ \Rightarrow \sin{5\theta} & = \Im (\color{#3D99F6}{\cos{\theta}} + i \color{#D61F06}{\sin{\theta}})^5 \quad \quad [\text{Let } \color{#3D99F6} {a=\cos{\theta}} \text{ and } \color{#D61F06} {b=\sin{\theta}} ] \\ & = \Im (\color{#3D99F6}{a} + i \color{#D61F06}{b})^5 \\ & = \Im (a^5 + i5a^4b - 10a^3b^2 - i10a^2b^3 + 5ab^4 + ib^5) \\ & = 5a^4b - 10a^2b^3 + b^5 \\ & = 5(1-b^2)^2b - 10(1-b^2)b^3 + b^5 \\ & = 5(1-2b^2+b^4)b - 10b^3 + 10 b^5 + b^5 \\ & = 5b -10b^3+5b^5 - 10b^3 + 10 b^5 + b^5 \\ & = 16b^5 - 20b^3 + 5b \\ & = 16\sin^5{\theta} - 20\sin^3{\theta} + 5\sin{\theta} \end{aligned}

a + b + c 14 = 16 + 20 + 5 14 = 27 \Rightarrow a + b + c -14 = 16+20+5-14 = \boxed{27}

Moderator note:

Great usage of the Euler identity to convert multiple angles into power form.

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