Finding angles in radians from range -pi < alpha < 0

D

deeeelg

New member
Joined
May 4, 2022
Messages
7
I am struggling to get my head around the following problem. I have had multiple people give me answers and most vary from each others but none could explain in a way i understood. the following question in relation to the range above

cos alpha = sqrt 3/2

is it easier to understand from the unit circle or the cosine wave?
 
Last edited:
I would draw the angle in the 4th quadrant.
There is a special right triangle whose hypotenuse is 2, one leg is sqrt(3) and the other leg is ____.
You should know everything about this triangle including the value for <a.

If you can't figure out what a equals, then post back with your labelled diagram.
 
This is where i currently am with this question. I am not confident on my findings though
 

Attachments

  • q4.jpg
    q4.jpg
    2.3 MB · Views: 7
deeeelg said:
I am struggling to get my head around the following problem. I have had multiple people give me answers and most vary from each others but none could explain in a way i understood. the following question in relation to the range above

cos alpha = sqrt 3/2

is it easier to understand from the unit circle or the cosine wave?
Click to expand...
Further to Steven G's advice (above) it may help you to copy into your workbook (then fill in all the 'answers' on it) the worksheet (below) that I give out to my pupils to complete when teaching them about Trigonometry.

Once you have completed the worksheet you should be in a much better position to identify any angle (and convert it to Radians) that complies with the given condition in the stated range. ?

Exact Values Worksheet:-

Exact-Values.png
​
 
Last edited:
deeeelg said:
cos alpha = sqrt 3/2

is it easier to understand from the unit circle or the cosine wave?
Click to expand...
PS: To answer your specific question: -
Personally, I would think it is "easier to understand from the unit circle" but it is easiest to understand when you are simply aware of what angle has a Cosine Ratio of "sqrt 3/2" (and why).
Hence the advice you've been given above.
NB: This is all assuming that "sqrt 3/2" means: \(\displaystyle \frac{\sqrt{3}}{2}\) and not \(\displaystyle \sqrt{\frac{3}{2}}\)
 
deeeelg said:
This is where i currently am with this question. I am not confident on my findings though
Click to expand...
For some inexplicable reason, your (2nd) post did not appear (for me at least) until after both of my posts!!
Now I have seen it, I am at a loss to know what your problem is?
Which of those angles are in the range: -π < alpha < 0 ? (Q.E.D.)
 
Essentially this is the snippet of the revision we have. Yet i am struggling. It is the range part that i cant figure out - or am i reading into it too much.

1651798673221.png
 
The question states \(\displaystyle -\pi<\alpha<0\).
This means two things:
1. The answer is required in radians
2. The answer must lie between \(\displaystyle -\pi \) and \(\displaystyle 0\).

Your diagram shows answers in the first and fourth quadrant. Good starting point BUT neither \(\displaystyle \frac{\pi}{6}\) nor \(\displaystyle \frac{11\pi}{6}\) are in the required domain.

The region where \(\displaystyle -\pi<\alpha<0\) begins at the pos x-axis (0) and swings around clockwise to the neg x-axis (\(\displaystyle -\pi\)).

If you do this, you can see that the only answer is in the fourth quadrant (at the position you have labelled \(\displaystyle \frac{11\pi}{6}\)).

BUT remember about coterminal angles. Coterminal angles "terminate" or end up at the same position.

So \(\displaystyle \frac{11\pi}{6}\) is coterminal with \(\displaystyle \frac{-\pi}{6}\). You should be able to see this from your diagram.

\(\displaystyle \frac{-\pi}{6}\) is now in the required domain since it is between \(\displaystyle -\pi \) and \(\displaystyle 0\).
 
deeeelg said:
Essentially this is the snippet of the revision we have. Yet i am struggling. It is the range part that i cant figure out - or am i reading into it too much.

View attachment 32604
Click to expand...
This is how I approach these problems.

1. Learn the sin, cos and tan of 30, 60 and 45 degrees (see post #4) or at least remember how to construct the relevant triangles and work them out.
2. Forget (for the moment) any negative signs in the question and work out the basic acute angle involved. (Eg Q3. What is the acute angle whose sin is 1/2 ?).
3. Consider the domain you are given. (\(\displaystyle -\pi<\alpa<0\) means clockwise from 0 to \(\displaystyle \pi\).
4. Draw a diagram and identify the quadrant/s in which the solution lies. (Do you know the ASTC rule?) This is where you have to take into account the sign (+/-).
5. Swing your acute angle into that quadrant from the horizontal axis and name it to suit the domain.
 
deeeelg said:
Essentially this is the snippet of the revision we have. Yet i am struggling. It is the range part that i cant figure out - or am i reading into it too much.

View attachment 32604
Click to expand...
From both your OP and your 2nd post (when I eventually saw it!) I was under the impression that you were only required to consider the angle(s) where cos alpha = \(\displaystyle \frac{\sqrt{3}}{2}\) !

This is an(other) example which strongly reinforces the need for you to provide details of your whole problem in your OP! ? (Please read the Posting Guidelines before posting: "2. Post the exercise or your question completely and accurately."!)

However, others (esp. @Harry_the_cat & @Subhotosh Khan ?) have now provided you with full information to complete your task.
 
You must log in or register to reply here.
Top