# Thread: 2 very difficult trig problems

1. ## 2 very difficult trig problems

Having an extremely difficult time solving 2 problems. Have posted it as a photo attachment. The first 2 paragraphs & diagram relate to the 1st problem. The last paragraph is for the second problem.

For the second problem, I tried using double-angle formula, but I don't know where to go after that. For the first problem, I am completely lost. Any help would be appreciated. Thanks

2. ## Re: 2 very difficult trig problems

I don't understand why changing the number of bearings changes anything. It just looks like it would fill up more of the space around the hub. If you could explain that I might be able to help.

3. ## Re: 2 very difficult trig problems

...And in the future please put your problem so it's right side up!

-Dan

4. ## Re: 2 very difficult trig problems

I think the diagram is a little misleading. It's just meant to show you what a, b, and c are. When there are three bearings, they fill up the entire circle, touching each other (i.e. a will be very small and b very big). As you add more ball bearings, a increases and b decreases. I think that's what the problem is trying to say.

5. ## Re: 2 very difficult trig problems

The attached figure explains the first problem more clearly. Fig 1 to 4 depicts the configuration for 3,4,5,6 ball bearing cases receptively.

As in Fig 5 lets consider the n-ball bearing case.

$\sin \theta = \frac{ID}{DL} = \frac{b}{a+b} \implies \frac{a}{b} = \csc \theta - 1$ where $\theta = \frac{\pi}{n}$ as it depicts a regular n-sided polygon.

Other ratios are easy as we have $c= a+2b$

Kalyan.

6. ## Re: 2 very difficult trig problems

HINT: For the second problem use the identity $\sin (90 \pm x) = \mp \cos x$ and the identity $sin^2 x + cos^2 x =1$.

Kalyan.

9. ## Re: 2 very difficult trig problems

Originally Posted by ibdutt
I have gone wrong and i am sorry. kalyanram is very correct and MINOANMAN has referred a good site.