Pre-Test on Trigonometric Equations and Applications

1. May 14, 2005

Lucretius

I have a math test on the chapter on Tuesday, and my teacher handed out the pre-test on Thursday. There are a few problems I am totally stumped on, and figured the math geniuses here could give me some help. Some I can get somewhere with, some I don't know where to begin.

Here is one problem I am stuck on:

At CI State Park, the average height of the tide in feet t hours after hight tide is given by the function

$$h(t)=2\cos\left(\frac{2\pi}{13}\left(t-2\right)\right)+4$$

a. What is the smallest positive value of t corresponding to the low tide?

The answer to a is 8:30. How did they come up with this answer?

2. May 14, 2005

dextercioby

Hmm,i dunno,calculus,maybe...?Or even simpler,what's the minimum value of the function (HINT:it has to do with the inferior boundedness of the cosine).....?

Daniel.

3. May 15, 2005

primarygun

It's better to express t as 8.5 rather than 8:30 I think.

4. May 15, 2005

dextercioby

Incidentally,he'll get the result as 8.5...If he wants to make the conversion,or not,that's another issue.

Daniel.

5. May 15, 2005

Lucretius

Here's a problem I got somewhere with, but I still get the wrong answer.

$$\frac{\tan\theta+\cot\theta}{\sec^2\theta}$$

I have to simplify this, and the answer ends up being \cot\theta

I did the following:

$$\frac{\tan\theta+\cot\theta}{\sec^2\theta}$$

$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ , $$\cot\theta = \frac{cos\theta}{\sin\theta}$$

So $$\frac{\sin\theta}{\cos\theta} + \frac{cos\theta}{\sin\theta} = \frac{sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$$ • $$\frac{\cos^2\theta}{1} (\sec^2})$$ which was our denominator to begin with

I get stuck here.

6. May 15, 2005

dextercioby

Nope,it's

$$\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}\frac{\cos^{2}\theta}{1}=\frac{1}{\sin\theta\cos\theta}\cdot\cos^{2}\theta=\frac{\cos\theta}{\sin\theta}=\cot\theta$$

Q.e.d.

Daniel.

7. May 15, 2005

Lucretius

Oh yeah

$$\sin^2\theta+\cos^2\theta=1$$

8. May 16, 2005

Lucretius

All right, the test is tomorrow and one or two of these problems are still giving me trouble.

$$\sin(x)+\sqrt{3}\cos(x)=0$$

So I did the following: took the square of everything to get rid of the square root, and I got:

$$\sin^2(x)+3\cos(x)=0$$

Then I changed $$\sin^2(x)$$ to be $$1-\cos^2(x)$$ and got:

$$1-\cos^2(x)+3\cos(x)=0$$

After I rearranged it, I got:

$$\cos^2(x)-3\cos(x)+1$$

I have tried breaking this down into something like:

$$\left(\cos+1\right)\left(\cos-1\right)$$

but it does not seem to work. I am supposed to get the values for $\cos$ that allow me to get as answers $2\pi/3$ and $5\pi/3$

Where did I go wrong?

9. May 16, 2005

what

What's the lowest value the cosine function could possibly have?

10. May 16, 2005

Gale

your first step is wrong, you can't square the terms individually to get rid of the root(3).

try just looking at it like this:
$$sinx= -\sqrt{3}cosx$$
then think of the unit circle values that would let this be true.

11. May 16, 2005

Lucretius

Ah, gale thanks, that helped a lot. If I can't solve another problem I'll head it your way, but I don't think I will, so just wish me luck on my test tomorrow everyone!

12. May 16, 2005

what

How about cos has a lowest value of -1 and that is when $$x=\pi$$. Therfore, $$\frac{2\pi}{13}\left(t-2\right)= \pi$$

13. May 16, 2005

Curious3141

Even easier :

Let $t = \tan\theta$

$$\frac{t + \frac{1}{t}}{1 + t^2}$$
$$=\frac{1+t^2}{t(1+t^2)}$$
$$=\frac{1}{t} = \cot\theta$$

14. May 17, 2005

dextercioby

It would help if here were to divide through $\cos x$ and then solve this

$$\tan x=-\sqrt{3}$$

in the reals...

Daniel.