- #1

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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.

cos(60)° = ½?

Thanks.

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- Thread starter pairofstrings
- Start date

- #1

- 385

- 6

- Summary:
- How do I know that cosine of 1 degree is: 0.9998

Why is cos (1)° = 0.9998?

cos(60)° = ½?

Thanks.

cos(60)° = ½?

Thanks.

- #2

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How is the cosine defined?Summary::How do I know that cosine of 1 degree is: 0.9998

Why is cos (1)° = 0.9998?

cos(60)° = ½?

Thanks.

- #3

FactChecker

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- #4

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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!

- #5

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Have you tried doing a bit of trigonometry using the fact that 30 degrees is a third of a right angle?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

- #6

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It's hard to understand the statement.Have you tried doing a bit of trigonometry using the fact that 30 degrees is a third of a right angle?

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

FactChecker

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Typically, the ##1\sqrt 3## is not considered a radius, the 2 is the radius.The radius of unit circle is 1 unit that is opposite to 30° angle. I don't see how 1√3 is a 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

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There must be an elementary proof using the trigonometry of a unit square or a circle.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.

I must admit I didn't know exactly how to.do this. And, I haven't found anything easy. Yet.

- #9

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- #10

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Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$

- #11

- 385

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The graphs don't match.Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$

- #12

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What does that mean?The graphs don't match.

- #13

FactChecker

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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.

- #14

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Prove it!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.

- #15

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What does that mean?

I tried to plot the graphs of $$\cos 30, \sin 60, 2\sin 30 \cos 30$$ $$\cos30$$ is not same as $$\sin60$$ and $$2\sin30\cos30$$Algebra to the rescue! $$\cos 30 = \sin 60 =2\sin 30 \cos 30$$

- #16

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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. :-)Prove it!

- #17

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Well, they are all the same.I tried to plot the graphs of $$\cos 30, \sin 60, 2\sin 30 \cos 30$$ $$\cos30$$ is not same as $$\sin60$$ and $$2\sin30\cos30$$

Use a calculator if you don't believe me.

- #18

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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.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. :-)

- #19

Mark44

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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.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##

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

pasmith

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Take an equilateral triangle of side 2. Cut it in half along a bisector.

- #21

pbuk

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Yes, you are obsessed with plotting graphs, all your questions are about plotting graphs but you cannot learn anything this way.Plotting graphs doesn't make any sense, because all of these expressions just represent numbers.

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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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.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.

- #23

pbuk

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But it continued withThe problem started with why are those particular side lengths are determined by the 30 degree and 60 degree angles.

So while Pythagoras may not be sufficient he is necessary.In a right-angled triangle ... Where did the side lengths 1, 1√3, 2 come from?

Last edited:

- #24

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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).But it continued with

So while Pythagoras may not be sufficient he is necessary.

- #25

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How did I not see that?Take an equilateral triangle of side 2. Cut it in half along a bisector.

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