Cosine of 1 degree and cosine of 60 degrees?

In summary, the adjacent side of a 30-60-90 right triangle is given as 1√3 units, the hypotenuse is 2 units, and the opposite side is 1 unit. This can be derived from the cosine and sine values of 30 and 60 degrees, which can be found using trigonometry or a calculator. It is also related to the side lengths of an equilateral triangle. However, a deeper understanding of trigonometry is necessary to fully comprehend these concepts.
  • #1
TL;DR 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
pairofstrings said:
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
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
pairofstrings said:
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
PeroK said:
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
pairofstrings said:
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
pairofstrings said:
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
Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$
 
  • #11
PeroK said:
Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$
The graphs don't match.
 
  • #12
pairofstrings said:
The graphs don't match.
What does that mean?
 
  • #13
pairofstrings said:
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
FactChecker said:
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
PeroK said:
What does that mean?
PeroK said:
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 $$2\sin30\cos30$$
 
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  • #16
PeroK said:
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
pairofstrings said:
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
FactChecker said:
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
pairofstrings said:
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.
 
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  • #20
pairofstrings said:
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
Mark44 said:
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
pbuk said:
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
FactChecker said:
The problem started with why are those particular side lengths are determined by the 30 degree and 60 degree angles.
But it continued with
pairofstrings said:
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
pbuk said:
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
pasmith said:
Take an equilateral triangle of side 2. Cut it in half along a bisector.
How did I not see that?
 
  • #26
FactChecker said:
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).
As I wrote in post #19, just before @pasmith's post, the problem can be done without invoking trig -- just with a bit of geometry and Pythagorus. Equilateral triangles are studied in geometry, along with the facts that all three sides and angles are equal -- at least they were when I had geometry in the last century.
 
  • #27
The OP also asked about the cosine of one degree. The easiest way I see to get that is that for small angles, ## \sin x \approx x ## when ## x ## is measured in radians, and ## \cos^2 x=1-\sin^2 x ## for all ## x ##. In more detail, ## \sin x=x-x^3/3!+... ## for ## x ## measured in radians.
 
  • #28
pairofstrings said:
Summary:: How do I know that cosine of 1 degree is: 0.9998

Why is cos (1)° = 0.9998?
cos(60)° = ½?
Thanks.
For cos of 1 deg, use power series expansion ##cos(x) \approx 1-\frac{x^2}{2}## where ##x=\frac{\pi}{180}##.
 
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  • #29
mathman said:
For cos of 1 deg, use power series expansion ##cos(x) \approx 1-\frac{x^2}{2}## where ##x=\frac{\pi}{180}##.
At 1 degree, the small angle approximation ##\sin {x} \approx x ## makes the estimate ##\cos {x} \approx \sqrt {1-x^2}## accurate to 7 digits.
EDIT: But the OP asks how we KNOW that cos(1 degree) = 0.9998. That requires that we can limit the error of the approximation, which complicates things.
 
Last edited:
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  • #30
pairofstrings said:
Summary:: How do I know that cosine of 1 degree is: 0.9998

Why is cos (1)° = 0.9998?
cos(60)° = ½?
Thanks.
That for 1 degree, just draw the figure and your guess should be about what you have there. For the 60 degree, this is already obvious through fairly simple Geometry.
 
  • #31
pbuk said:
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
So this points back to some of the simpler Geometry you study in high school, about right triangles, some regular polygons, equilateral triangles.
 
  • #32
pasmith said:
Take an equilateral triangle of side 2. Cut it in half along a bisector.
I think you really meant "sides o three"? Two of the sides become hypotenuses of the two new triangles after bisecting the equilat. into the two right triangles.
 
  • #33
pasmith said:
Take an equilateral triangle of side 2. Cut it in half along a bisector.
symbolipoint said:
I think you really meant "sides o three"? Two of the sides become hypotenuses of the two new triangles after bisecting the equilat. into the two right triangles.
No, @pasmith meant what he wrote; namely, an equilateral triangle whose sides are of length 2.
triangle.png


The angle at the lower left is 60°, so ##\cos(60°) = \frac 1 2##.
 
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  • #34
pairofstrings said:
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?
Pytagoras's theorem
[tex]a^2+b^2=c^2[/tex]
says for this case
[tex]1^2+\sqrt{3}^2=2^2[/tex]
You can easily find proof of Pytagoras's theorem on Web.

You can make a equilateral triangle from the two of these triangles then you will find the angles are 30- 60- 90 as shown in post #33.
 
  • #35
pairofstrings said:
Why is cos (1)° = 0.9998?
If you have got understood the formula
[tex]\cos2\theta=2cos^2\theta-1[/tex]
we can make use of it sequentially to get half-angle, quarter angle, one eighth angle,... of cosine.
The proof of the formula is easily found in the web.

Starting from cos 60 degree = 1/2, cos30, cos15, cos(15/2), cos(15/4),cos(15/8), cos(15/16)

Thus we can calculate value of cos (15/16) degree and may hope it an approximate value of cos 1 degree. Please try it if you have an interest in this method.
 

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