Cosine of 1 degree and cosine of 60 degrees?

  • #1
Summary:
How do I know that cosine of 1 degree is: 0.9998
Why is cos (1)° = 0.9998?
cos(60)° = ½?
Thanks.
 

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  • #2
PeroK
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Summary:: How do I know that cosine of 1 degree is: 0.9998

Why is cos (1)° = 0.9998?
cos(60)° = ½?
Thanks.
How is the cosine defined?
 
  • #3
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Have you taken a class that included trigonometry yet? That is where trigonometric functions like sin(), cos(), tan(), etc. are studied.
 
  • #4
In a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit?
Where did the side lengths 1, 1√3, 2 come from? If it is from Unit Circle then why adjacent side is 1√3. The radius of unit circle is 1 unit that is opposite to 30° angle. I don't see how 1√3 is a radius; there is only one radius in one unit circle!
body_trig-300x168.png
 
  • #5
PeroK
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In a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit?
Where did the side lengths 1, 1√3, 2 come from? If it is from Unit Circle then why adjacent side is 1√3. The radius of unit circle is 1 unit that is opposite to 30° angle. I don't see how 1√3 is a radius; there is only one radius in one unit circle!
View attachment 288065
Have you tried doing a bit of trigonometry using the fact that 30 degrees is a third of a right angle?
 
  • #6
Have you tried doing a bit of trigonometry using the fact that 30 degrees is a third of a right angle?
It's hard to understand the statement.
One thing I know is side length is minimum if the opposite angle is minimum,
Side length is maximum if the opposite angle is maximum.
Side length is moderate if the opposite angle is between max and min.
 
  • #7
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The radius of unit circle is 1 unit that is opposite to 30° angle. I don't see how 1√3 is a radius;
Typically, the ##1\sqrt 3## is not considered a radius, the 2 is the radius.
In general, cos() is just looked up in a table or found with a calculator. There are formulas that can be applied for the sums of two angles and other special situations.
The cos() of a few angles (0, 30, 45, 60, 90) are expected to be known. (A memory trick is that they follow a pattern ##\sqrt 4/2, \sqrt 3/2, \sqrt 2/2, \sqrt 1/2, \sqrt 0/2##)
Are you currently taking a trigonometry class? That is where these subjects are taught.
 
  • #8
PeroK
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It's hard to understand the statement.
One thing I know is side length is minimum if the opposite angle is minimum,
Side length is maximum if the opposite angle is maximum.
Side length is moderate if the opposite angle is between max and min.
There must be an elementary proof using the trigonometry of a unit square or a circle.

I must admit I didn't know exactly how to.do this. And, I haven't found anything easy. Yet.
 
  • #9
All I am trying to understand is if there is a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit? Where did they come from?
 
  • #10
PeroK
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Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$
 
  • #11
Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$
The graphs don't match.
 
  • #13
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All I am trying to understand is if there is a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit? Where did they come from?
In a triangle with hypotenuse=2, the adjacent side length is 2*cos(30deg)=2*##\sqrt 3/2##=##\sqrt 3##. The opposite side is 2*sin(30deg)=2*1/2=1.
 
  • #14
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In a triangle with hypotenuse=2, the adjacent side length is 2*cos(30deg)=2*##\sqrt 3/2##=##\sqrt 3##. The opposite side is 2*sin(30deg)=2*1/2=1.
Prove it!
 
  • #15
What does that mean?
Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$
I tried to plot the graphs of $$\cos 30, \sin 60, 2\sin 30 \cos 30$$ $$\cos30$$ is not same as $$\sin60$$ and $$2\sin30\cos30$$
 
  • #16
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Prove it!
Ha! I thought that the sum-angle formula would do it, but then I saw your post and realized that it may not be simple. I'll take your word for it. :-)
 
  • #17
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I tried to plot the graphs of $$\cos 30, \sin 60, 2\sin 30 \cos 30$$ $$\cos30$$ is not same as $$\sin60$$ and $$2\sin30\cos30$$
Well, they are all the same.

Use a calculator if you don't believe me.
 
  • #18
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Ha! I thought that the sum-angle formula would do it, but then I saw your post and realized that it may not be simple. I'll take your word for it. :-)
Once you get the sine and cosine for 30 degrees ( e.g. from the double angle formula), then you can return to the geometric approach to get them for 15 and 75 degrees.
 
  • #19
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I tried to plot the graphs of ##\cos 30, \sin 60, 2\sin 30 \cos 30## ##\cos30## is not same as ##\sin60## and ##2\sin30\cos30##
Plotting graphs doesn't make any sense, because all of these expressions just represent numbers. Furthermore, the various trig identities show that ##\cos(30°)## is exactly equal to ##\sin(60°)##, and the double angle formula (also an identity) shows that ##\sin(60°) = \sin(2 * 30°) = 2\sin(30°)\cos(30°)##. Also, if you're using a calculator, make sure it's in degree mode, not radian mode.

As for why the sides of a 30-60-90 right triangle are 1, 2, and ##\sqrt 3##, draw an equilateral triangle whose sides are 2 units. The interior angles of this triangle are all 60 deg. Draw an altitude from the middle of one side to the opposite vertex. Each half is a 30-60-90 right triangle. The hypotenuse is 2 units, the short side is 1 unit, and you can use the Theorem of Pythagoras to find the third side.

If you haven't studied trigonometry yet, you're not going to get very far with the kinds of questions you're asking.
 
  • #20
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All I am trying to understand is if there is a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit? Where did they come from?

Take an equilateral triangle of side 2. Cut it in half along a bisector.
 
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  • #21
pbuk
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Plotting graphs doesn't make any sense, because all of these expressions just represent numbers.
Yes, you are obsessed with plotting graphs, all your questions are about plotting graphs but you cannot learn anything this way.

If you don't understand why the base of that triangle is ##\sqrt{3} ## long then you are not ready for sine and cosine yet, you need to start with Pythagoras. This course should do:
https://www.khanacademy.org/math/geometry/hs-geo-trig
 
  • #22
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If you don't understand why the base of that triangle is ##\sqrt{3} ## long then you are not ready for sine and cosine yet, you need to start with Pythagoras.
The problem started with why are those particular side lengths are determined by the 30 degree and 60 degree angles. So Pythagoras is not enough.
 
  • #23
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The problem started with why are those particular side lengths are determined by the 30 degree and 60 degree angles.
But it continued with
In a right-angled triangle ... Where did the side lengths 1, 1√3, 2 come from?
So while Pythagoras may not be sufficient he is necessary.
 
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  • #24
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But it continued with

So while Pythagoras may not be sufficient he is necessary.
I would get the two side lengths using trigonometry, not using the Pythagorian Theorem at all (although it could be used to get the third side).
 
  • #25
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Take an equilateral triangle of side 2. Cut it in half along a bisector.
How did I not see that?
 

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