MHB Trigonometric Identity Correction: Solving a Complex Equation

AI Thread Summary
The discussion centers on a trigonometric identity involving the equation $\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}$. A misprint is identified in the original identity, which should correctly state $d^3(4c^2-a^2)=c^4(3d-b)$. Participants express gratitude to Opalg for pointing out the error. The correct formulation is essential for solving the complex equation accurately. Clarifying this identity is vital for further mathematical discussions and solutions.
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If $\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}$, show that $2d^3(2c^3-a^2)=c^4(3d-b)$.
 
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anemone said:
If $\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}$, show that $2d^3(2c^3-a^2)=c^4(3d-b)$.

I'm sorry, folks!(Tmi)(Worried) But there's a misprint in the identity because it should read as $d^3(4c^2-a^2)=c^4(3d-b)$.:o

I must thank Opalg for letting me know something doesn't look right with the initial identity!(Sun)
 
anemone said:
I'm sorry, folks!(Tmi)(Worried) But there's a misprint in the identity because it should read as $d^3(4c^2-a^2)=c^4(3d-b)$.:o

Let
$$\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}=\lambda$$

$$\sin 3x=3\sin x-4\sin^3 x \Rightarrow b\lambda =3d\lambda-4d^3\lambda^3 \Rightarrow \lambda^2=\frac{3d-b}{4d^3}$$

Also,
$$\sin 4x=2\sin 2x \cos 2x \Rightarrow 2c\lambda \sqrt{1-c^2\lambda^2}=a\lambda \Rightarrow \lambda^2=\frac{4c^2-a^2}{4c^4}$$

Equating the two expression for $\lambda^2$,
$$\frac{3d-b}{4d^3}=\frac{4c^2-a^2}{4c^4}$$
$$\Rightarrow d^3(4c^2-a^2)=c^4(3d-b)$$

$\blacksquare$
 
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