MHB Is the Bottom Equality in this Complex Mathematical Equation Correct?

AI Thread Summary
The discussion centers on verifying the correctness of a complex mathematical equation involving derivatives and trigonometric functions. The original poster expresses uncertainty after extensive review and seeks confirmation from others. A participant confirms that the bottom equality of the equation is indeed correct. The equation includes terms with coefficients influenced by constants A, B, and their derivatives, as well as trigonometric identities. The conversation concludes with validation of the equation's accuracy.
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I have another answer to this but I believe this one is correct. I need someone else to check it out since I have been looking at it too long. Is the bottom equality correct?

\begin{alignat*}{3}
\frac{\partial^2}{\partial t^2}x_1 + x_1 & = & F\cos t - 2[-A'\sin t + B'\cos t] - c[-A\sin t + B\cos t] - (A\cos t + B\sin t)^3\\
& = & F\cos t + 2A'\sin t - 2B'\cos t + cA\sin t - cB\cos t - A^3\cos^3 t - 3A^2B\cos^2 t\sin t\\
& & - 3AB^2\cos t\sin^2 t - B^3\sin^3 t\\
& = & \cos t\left(F - 2B' - cB - \frac{3}{4}AB^2 - \frac{3}{4}A^3\right) + \sin t\left(2A' + cA - \frac{3}{4}A^2B - \frac{3}{4}B^3 \right)\\
& & + \left(\frac{3}{4}AB^2 - \frac{1}{4}A^3\right)\cos 3t - \left(\frac{3}{4}A^2B - \frac{1}{4}B^3\right)\sin 3t
\end{alignat*}
 
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dwsmith said:
I have another answer to this but I believe this one is correct. I need someone else to check it out since I have been looking at it too long. Is the bottom equality correct?

\begin{alignat*}{3}
\frac{\partial^2}{\partial t^2}x_1 + x_1 & = & F\cos t - 2[-A'\sin t + B'\cos t] - c[-A\sin t + B\cos t] - (A\cos t + B\sin t)^3\\
& = & F\cos t + 2A'\sin t - 2B'\cos t + cA\sin t - cB\cos t - A^3\cos^3 t - 3A^2B\cos^2 t\sin t\\
& & - 3AB^2\cos t\sin^2 t - B^3\sin^3 t\\
& = & \cos t\left(F - 2B' - cB - \frac{3}{4}AB^2 - \frac{3}{4}A^3\right) + \sin t\left(2A' + cA - \frac{3}{4}A^2B - \frac{3}{4}B^3 \right)\\
& & + \left(\frac{3}{4}AB^2 - \frac{1}{4}A^3\right)\cos 3t - \left(\frac{3}{4}A^2B - \frac{1}{4}B^3\right)\sin 3t
\end{alignat*}

Yes it's correct. (Yes)
 
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