Solving Trig Problems: Evaluate cos 300° and sin(-120°)

  • Thread starter TheKracken
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In summary, the student is struggling with remembering the trig functions and is not consistent with how to write the equations.
  • #1
356
7

Homework Statement



Evaluate the following exactly.

cos 300°
sin(-120°)

Homework Equations


Unit circle


The Attempt at a Solution


I seem to have a problem with the remembering if it is going to be the position on the unit circle such as in these cases I thought it was
cos 300°= 5∏/3 but it is actually I think 1/2 so I got the problem wrong
then you have
sin(-120°) = 4∏/3 but this is wrong and I think it is -√3 / 2 so yeah then I got this one right
csc120°= 2√3 / 3
I just never know when I am supposed to use which one for the exact value. Could someone help out please?
 
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  • #2
TheKracken said:

Homework Statement



Evaluate the following exactly.

cos 300°
sin(-120°)

Homework Equations


Unit circle


The Attempt at a Solution


I seem to have a problem with the remembering if it is going to be the position on the unit circle such as in these cases I thought it was
cos 300°= 5∏/3 but it is actually I think 1/2 so I got the problem wrong
You're really garbling up a bunch of different stuff. cos 300° ≠ 5∏/3, but 300° = 5∏/3. The reference angle for 5∏/3 is -∏/3, and this angle has the same cosine as ∏/3.

Writing cos 300°= 5∏/3 is incorrect, but cos(300°)= cos(5∏/3) would be correct. Be mindful that the cosine of something is not the same as the something.
TheKracken said:
then you have
sin(-120°) = 4∏/3 but this is wrong and I think it is -√3 / 2 so yeah then I got this one right
csc120°= 2√3 / 3
I just never know when I am supposed to use which one for the exact value. Could someone help out please?
 
  • #3
So then why is arcsin(-1/2) = -Pi/6 Is the inverse just going back to the reference angle then?
 
  • #4
Yes, because the trig functions are periodic, they are not "one to one" and so do not have true "inverses". In order to talk about inverse for the trig functions, we have to restrict their domains. The standard convention for "sine" is to restrict to between [itex]-\pi/2[/itex] (-90 degrees) and [itex]\pi/2[/itex] (90 degrees). If x is positive, [itex]sin^{-1}(x)[/itex] is between 0 and [itex]\pi/2[/itex]. If x is negative, [itex]sin^{-1}(x)[/itex] is between [itex]-\pi/2[/itex] and 0.

Personally, for "sin(300)" I would note that 300= 360- 60 and use "sin(a-b)= sin(a)cos(b)- cos(a)sin(b)" so that sin(300)= sin(360)cos(60)- cos(360)sin(60)=0- sin(60).
 

1. What is the process for solving trigonometry problems?

The process for solving trigonometry problems involves using the given information to determine which trigonometric function and formula to use, substituting the given values into the formula, and then using algebraic manipulation to solve for the unknown variable.

2. How do you evaluate cos 300°?

To evaluate cos 300°, we can use the unit circle or apply the cosine function using the reference angle of 60° in the fourth quadrant. This yields a value of -0.5 for cos 300°.

3. What is the difference between evaluating cos 300° and cos -300°?

The difference between evaluating cos 300° and cos -300° is that cos 300° is in the fourth quadrant, while cos -300° is in the second quadrant. This results in different reference angles and therefore different values for the cosine function.

4. How do you evaluate sin(-120°)?

To evaluate sin(-120°), we can use the unit circle or apply the sine function using the reference angle of 60° in the third quadrant. This yields a value of -√3/2 for sin(-120°).

5. Can trigonometry problems have multiple solutions?

Yes, trigonometry problems can have multiple solutions. This is because there are multiple points on the unit circle with the same sine, cosine, or tangent value. Additionally, some problems may have infinite solutions, such as when solving for an angle using an inverse trigonometric function.

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