Pre-Test on Trigonometric Equations and Applications

[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 height 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?

 

[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.

 

[primarygun]

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

 

[dextercioby]

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

Daniel.

 

[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.

 

[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.

 

[Lucretius]

Oh yeah

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

 

[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?

 

[what]

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

 

[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.

 

[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!

 

[what]

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

 

[Curious3141]

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.

Even easier :

Let $t = \tan\theta$

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

 

[dextercioby]

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:
$$\sin x= -\sqrt{3}\cos x$$
then think of the unit circle values that would let this be true.

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

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

in the reals...

Daniel.