Find all angles α in radians from the range −π < α < 0 for cos α = √3/2

[HiMyNameIs]

Find all angles α in radians from the range −π < α < 0 for cos α = √3/2

Disclosure; this is for an assignment but I'm on struggle street and It doesn't seem to be clicking in my head and making sense.

Q. Find all angles α in radians from the range −π < α < 0 such that:
a) cos α = √3/2

This is the working out I know how to do, but I don't know if what I'm doing is correct.


arccos(√3/2) = pi/6 = 0.5236
arccos(0.5236) = 58.42^o
180^o - 58.42^o = 121.58^o
cos(58.42) = 0.5236
sin(58.42) = 0.8519
cos(121.58) = -0.5236
sin(121.58) = 0.8519
When it says find all angles between pi and 0, what exactly does that mean? have I found all the angles in radians or do I need to convert 58.42^o and 121.58^o to radians?

58.42 x pi/180 = 1.019 radians?

 

[stapel]

Q. Find all angles α in radians from the range −π < α < 0 such that:
a) cos α = √3/2

This is the working out I know how to do, but I don't know if what I'm doing is correct.

arccos(√3/2) = pi/6 = 0.5236

Why use decimal approximations (that is, inexact "degree" measures)? Stick to the exact (radian) form:

. . . . .\(\displaystyle \cos(\alpha)\, =\, \dfrac{\sqrt{\strut 3\,}}{2}\, \mbox{ for }\, \alpha\, =\, \dfrac{\pi}{6}\)

Now look at the cosine wave. Since you're working with the interval from -pi to zero, let's use the period from -2pi to zero. You know that the cosine wave starts its period at 1, goes down to zero, crosses below the x-axis to -1, curves back up to zero, and then ends up at 1 again. You're looking only at that second half, where the curve goes from -1, up to zero, and then ends at 1.

You know that the y-values (the curve values) on this half-period are kind of symmetric; the values have opposite signs but equal absolute values as you come in from either end (from -pi/2 toward -pi/4, and from zero back toward -pi/4). Will there be any more than one location on the specified interval where the cosine will take on the specified positive value?

You also know that the cosine wave is symmetric about the y-axis; the values on the positive side mirror those on the negative side. You've found a solution on the interval zero to pi/4. What does this tell you about the solution on the interval -pi/4 to zero?

When it says find all angles between -pi and 0, what exactly does that mean?

It means exactly that: find all angles "alpha" for which cos(alpha) is the specified value, where "alpha" is between negative pi and zero.

 

[pka]

Disclosure; this is for an assignment but I'm on struggle street and It doesn't seem to be clicking in my head and making sense.

Q. Find all angles α in radians from the range −π < α < 0 such that:
a) cos α = √3/2

I have generations of students, even graduate students, who have used this half of an equilateral triangle, a \(\displaystyle \frac{\pi}{6},~\frac{\pi}{3},~\&~\frac{\pi}{2}\)

From this memory device, you can find well know angles at once.
The angles in quad IV have a positive cosines. So to answer to your question is \(\displaystyle -\dfrac{\pi}{6}\).

DON'T use a calculator with well-known angles. You can see that the above answer is so much cleaner then all that mess you posted.