Trigonometry equation, principal value(i dont get this at all)

1. Aug 31, 2008

joker2413

1.Solve the following for the principal values of the indicated variable.

2. 3cos x-1=2

3. 3cosx - 1 = 2
3cosx+1=0

im not sure what to do from here, I would ask a teacher or a tutor but it is summer work for a calculus course and i forgot what to do. I have looked in a precalculus book at the library and couldnt understand what i was supposed to do. Please explain how you solve this equation and I will be most grateful.

2. Aug 31, 2008

rock.freak667

Re-check this.

3cosx-1-2=0 gives what? (does -1-2=+1?)

3. Aug 31, 2008

joker2413

@_@ lol i must be tired... thanks

3cosX - 1 = 2
3cosX -3 = 0

but i still dont have any idea of what to do from there.

4. Aug 31, 2008

HallsofIvy

Staff Emeritus
3cos x- 1= 2
3cox x= 3
cos x= ?

5. Aug 31, 2008

joker2413

that would work to i just wasnt sure if there was any special rules with cos being multiplied..

so the final answer is cos x= 1? or would it be like... cos-1(cosx) = cos-1
X = cos-1

or do i have to do something else?

6. Aug 31, 2008

HallsofIvy

Staff Emeritus
cos(x)= 1 you should be able to remember almost without thinking!

But be very careful what you write. To me, cos-1 means cos(-1). I think you mean cos-1(1). What does your calculator say that is?

7. Aug 31, 2008

joker2413

yeah i did mean X= cos-1(1)

new here and didnt see the tags... sorry...

calc says cos-1(1) = 0 but i kind of remember cos x = 90º

which one is a principal value?

8. Aug 31, 2008

HallsofIvy

Staff Emeritus
Please, please, please, go back and review trigonometry! "cos x= 90º" doesn't even make sense: you take the cosine OF an angle, an angle is not the value of a cosine!

If you meant "cos(90º)= 0" that is also wrong. Now, exactly what definition of sine and cosine are you using? One problem with the old "trigonometry" definition of "opposite side over hypotenuse" and "near side over hypotenuse" is that your angles must larger than 0º and less than 90º in order to have a triangle: you can't have an angle in a triangle so that cos(x)= 1 because the near side and hypotenuse cannot be of the same length.

But in higher mathematics, we use sine and cosine as functions for many different applications that have nothing to do with triangles or even angles and we want to be able to define sin(x) and cosine(x) for any number x. A good way to do that is to use the unit circle. Draw a unit circle (center at (0,0), radius 1) on an xy- coordinate system. Starting from the point (1, 0), measure distance t counterclockwise around the circumference of the circle. The coordinates of the end point are, by definition, (cos(t), sin(t)). (If t is negative, measure clockwise.) That is, cos(t) and sin(t) are defined as the x and y coordinates of that point.

Notice that if you draw a line from that end point to the origin and also draw a line from that end point perpendicular to the x-axis, you make a right triangle having hypotenuse of length 1 (because the unit circle has radius 1), near side of length x and opposite side of length y so that cosine and sine of that angle are x/1= x and y/1= y.

But also notice that the "t" in cos(t) and sin(t) here is NOT an angle at all. It is a distance around the circle. If we measure the angle in radians, since radian angle measure is defined as the length of the arc subtending the angle defined by the length, with the unit circle the angle measure in radians is exactly the length of the arc that I am calling "t" here. Degree measure is convenient for working specifically with angles and triangles but for any other application of sine and cosine, we always use "radians".

Of course, since a unit circle with have circumference $2\pi r= 2\pi(1)= 2\pi$, we can also see that half a circle, corresponding to 180º, is $\pi$ radians and 1/4 of a circle, corresponding to 90º, is $\pi/2$ radians.

Now, on that unit circle graph you have drawn, mark the points (1, 0), (0, 1), (-1, 0), (0, -1). That should make it very easy to see what values of t give cos(t) and sin(t) equal to 0, 1, or -1.

9. Aug 31, 2008

joker2413

read what you said a few times but dont get some of it...

but for the problem would it be X = 0 and the principal values are X = $$\pi$$/2

and what im doing right now and for the past 5 days is reviewing trigonometry i just have a little trouble with some of it.