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Trigonometry factoring and identities

dwsmith

Well-known member
Feb 1, 2012
1,673
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*}
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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)