MHB When do the combined heights of two Ferris wheels reach 27.5m?

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The discussion focuses on determining the times when the combined heights of two Ferris wheels reach 27.5 meters, modeled by the equations F1 and F2. By equating the sum of the height functions to 27.5, the equation simplifies to a cosine equation involving a triple-angle identity. The solutions involve finding values of theta where cosine equals zero or ±√(1/2). The final expressions for time t are derived from these solutions, indicating the specific moments when the combined heights meet the specified requirement. The mathematical approach effectively combines trigonometric identities and algebraic manipulation to solve the problem.
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So I have figured out two equations model a Ferris wheel ride(two different ferris wheels).
F1 = Ferris wheel one
F2= Ferris wheel two

F1, $$h=-12\cos\frac{\pi}{10}t+12.5$$

F2, $$h=-12\cos\frac{\pi}{30}t+15$$

Now from these two equations, I want to know when F2+F1= 27.5m i.e I want to find the times when the combined height of each Ferris wheels seat adds up to 27.5m.
How would I go about doing this?
 
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If we equate sum of the two height functions to 27.5, we ultimately obtain:

$$\cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0$$

Let's let:

$$\theta=\frac{\pi}{30}t$$

And we may write:

$$\cos(3\theta)+\cos(\theta)=0$$

Applying a triple-angle identity for cosine, we ultimately get:

$$2\cos^3(\theta)-\cos(\theta)=0$$

Factor:

$$\cos(\theta)(2\cos^2(\theta)-1)=0$$

Can you proceed?
 
MarkFL said:
If we equate sum of the two height functions to 27.5, we ultimately obtain:

$$\cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0$$

Let's let:

$$\theta=\frac{\pi}{30}t$$

And we may write:

$$\cos(3\theta)+\cos(\theta)=0$$

Applying a triple-angle identity for cosine, we ultimately get:

$$2\cos^3(\theta)-\cos(\theta)=0$$

Factor:

$$\cos(\theta)(2\cos^2(\theta)-1)=0$$

Can you proceed?
\[ \cos \left(θ\right)=0\quad \mathrm{or}\quad \:2\cos ^2\left(θ\right)-1=0 \]

\[ \cos \left(θ\right)=0: θ=\frac{\pi }{2}+2\pi n,\:θ=\frac{3\pi }{2}+2\pi n \]

\[ 2\cos ^2\left(θ\right)-1=0: θ=\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=2\pi -\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=-\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n \] ?
 
Yes, for:

$$\cos(\theta)=0$$

This implies (where \(k\in\mathbb{Z}\)):

$$\theta=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2k+1)$$

$$t=\frac{30}{\pi}\theta=15(2k+1)$$

And for:

$$\cos(\theta)=\pm\frac{1}{\sqrt{2}}$$

This implies:

$$\theta=\frac{\pi}{4}+\frac{\pi}{2}k=\frac{\pi}{4}(2k+1)$$

$$t=\frac{30}{\pi}\theta=\frac{15}{2}(2k+1)$$
 
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