Pre-Test on Trigonometric Equations and Applications

In summary, the pre-test on Thursday gave the math geniuses at Daniel's school some problems to work on for the chapter test on Tuesday. One of the problems was expressed in terms of finding the lowest value of the cosine function when x=\pi. Daniel was able to solve this problem with the help of a simpler problem that had the same answer.
  • #1
Lucretius
152
0
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

[tex]h(t)=2\cos\left(\frac{2\pi}{13}\left(t-2\right)\right)+4[/tex]

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?
 
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  • #2
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
It's better to express t as 8.5 rather than 8:30 I think.
 
  • #4
Incidentally,he'll get the result as 8.5...:wink:If he wants to make the conversion,or not,that's another issue.

Daniel.
 
  • #5
Here's a problem I got somewhere with, but I still get the wrong answer.

[tex]\frac{\tan\theta+\cot\theta}{\sec^2\theta}[/tex]

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

I did the following:

[tex]\frac{\tan\theta+\cot\theta}{\sec^2\theta}[/tex]

[tex]\tan\theta = \frac{\sin\theta}{\cos\theta}[/tex] , [tex]\cot\theta = \frac{cos\theta}{\sin\theta}[/tex]

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

I get stuck here.
 
  • #6
Nope,it's

[tex] \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 [/tex]

Q.e.d.


Daniel.
 
  • #7
Oh yeah

[tex]\sin^2\theta+\cos^2\theta=1[/tex]
 
  • #8
All right, the test is tomorrow and one or two of these problems are still giving me trouble.

[tex]\sin(x)+\sqrt{3}\cos(x)=0[/tex]

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

[tex]\sin^2(x)+3\cos(x)=0[/tex]

Then I changed [tex]\sin^2(x)[/tex] to be [tex]1-\cos^2(x)[/tex] and got:

[tex]1-\cos^2(x)+3\cos(x)=0[/tex]

After I rearranged it, I got:

[tex]\cos^2(x)-3\cos(x)+1[/tex]

I have tried breaking this down into something like:

[tex]\left(\cos+1\right)\left(\cos-1\right)[/tex]

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

Where did I go wrong?
 
  • #9
What's the lowest value the cosine function could possibly have?
 
  • #10
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:
[tex] sinx= -\sqrt{3}cosx[/tex]
then think of the unit circle values that would let this be true.
 
  • #11
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
How about cos has a lowest value of -1 and that is when [tex]x=\pi[/tex]. Therfore, [tex]\frac{2\pi}{13}\left(t-2\right)= \pi[/tex]
 
  • #13
dextercioby said:
Nope,it's

[tex] \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 [/tex]

Q.e.d.


Daniel.

Even easier :

Let [itex]t = \tan\theta[/itex]

[tex]\frac{t + \frac{1}{t}}{1 + t^2}[/tex]
[tex]=\frac{1+t^2}{t(1+t^2)}[/tex]
[tex]=\frac{1}{t} = \cot\theta[/tex]
 
  • #14
Gale17 said:
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:
[tex] \sin x= -\sqrt{3}\cos x[/tex]
then think of the unit circle values that would let this be true.

It would help if here were to divide through [itex] \cos x [/itex] and then solve this

[tex] \tan x=-\sqrt{3} [/tex]

in the reals...

Daniel.
 

Related to Pre-Test on Trigonometric Equations and Applications

1. What are trigonometric equations?

Trigonometric equations are mathematical equations that involve trigonometric functions, such as sine, cosine, and tangent. These equations are used to solve for unknown angles or sides in a right triangle.

2. How do you solve a trigonometric equation?

To solve a trigonometric equation, you must first isolate the trigonometric function on one side of the equation. Then, you can use inverse trigonometric functions or trigonometric identities to solve for the variable.

3. What are the applications of trigonometric equations?

Trigonometric equations have many practical applications, such as calculating distances and heights in navigation and engineering, analyzing periodic phenomena in physics and astronomy, and modeling complex systems in mathematics and computer science.

4. What is the unit circle and how is it used in trigonometric equations?

The unit circle is a circle with a radius of 1 and centered at the origin on a Cartesian coordinate system. It is used in trigonometric equations to relate the coordinates of a point on the circle to the values of trigonometric functions. This allows for easy conversion between angles measured in degrees or radians.

5. How can I practice and improve my skills in solving trigonometric equations?

One way to practice and improve your skills in solving trigonometric equations is to do lots of practice problems. You can also use online resources, such as interactive quizzes and tutorials, or seek help from a tutor or teacher if you are struggling with a particular concept. Additionally, reviewing and understanding the fundamental principles and concepts of trigonometry can also help improve your skills in solving equations.

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