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 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?

 

[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.