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.
Dustinsfl
Messages
2,217
Reaction score
5
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*}
 
Mathematics news on Phys.org
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top