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**naturewild****Member**- Registered: 2005-12-04
- Posts: 30

How do I do it?

Like sin 90 = 1 and sin 210 = -1

Is there a chart that list the values and the only way for me is to memorize them?

Or is there some kind of pattern I can follow?

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

the 45, 45, 90 triangle and the 30,60,90 triangle can be used to find the sine cosine and tangents of multiples of 30 or 45.

There are some formula's which calculators use to calculate sine and cosine, but they are very very long and impracticle to do by hand. Better for computers or calculators.

Why can't you use a calculator? There are calculators to calculate sine and cosine that are small enough to fit in your mouth!

*Last edited by mikau (2006-09-11 14:29:28)*

A logarithm is just a misspelled algorithm.

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Here's a [hopefuly!] clear diagram of what mikau was talking about. On the left is an equilateral triangle with each side having a length of two, that is split in half to make two right-angled triangles. The height can be worked out from Pythagoras' theorem, the angles are known, and then the sine, cosine and tangent of 30[sup]o[/sup] and 60[sup]o[/sup] can be worked out.

On the right, take a right angled triangle with two sides of length one, the length of the other side can be worked out by Pythagoras, both angles other than the right angle will be 45[sup]o[/sup], and then sine cosine and tangent of 45[sup]o[/sup] can be worked out.

Bad speling makes me [sic]

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**naturewild****Member**- Registered: 2005-12-04
- Posts: 30

But how about something like cos 480?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

First off, you can take 360 off that to get cos 120, because you can always add or remove multiples of 360 from an angle without changing it.

Then you can refer to the CAST diagram and see that 120 is in the 2nd quadrant, so cos 120 is equivalent to -cos 60.

And you can use the 30-60-90 triangle to see that cos 60 is 0.5, which means that cos 480 is -0.5.

Why did the vector cross the road?

It wanted to be normal.

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**Scot****Guest**

Just use the unit circle. This is a circle centered at the origin with radius 1 (so it contains the points (1,0), (0,1), (-1,0), and (0, -1). Any angle you need can be found on this circle by moving around the circle in the counterclockwise direction. When you find the point on the circle that corresponds to the angle, a, you need, the point will be (x,y) where x=cos(a) and y=sin(a) and y/x = tan(a). The only thing you have to remember is the coordinates of these points, which takes some studying but it is definitely the best way to do find sin and cos once you know it.

**Ron****Guest**

Hey I just searched for this on google and I found this website.

Could I see that long and impracticable equation to figure out sin, cos, and tan? I've been looking everywhere, but I just can't find it.

I know all 3 are Ratios and I know about SOHCAHTOA and everything, but I just can't figure how to calculate it without using a calculator. Thanks in advance!

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

Do you mean the Taylor Series for sin, cos and tan? They are at:

http://en.wikipedia.org/wiki/Trigonometric_function#Series_definitions

(I should have a page on them, but don't yet)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

I have one way to figure them out without a calculator. But you need other tools.

Find a piece of hard-paper, draw a circle with radius of integer centimetres, and cut. Also depict the co-ords cross with pen. Next time you wanna figure out a sin, just measure it from the round paper you carry with an angle-measure and a ruler. Sure you can improve the special tool, too. You can paste co-ords paper onto it so you may not need a ruler, and you can draw 10°, 20°,... on to it so that you no longer need to carry the tool to measure the angle if you are not that exact about the result.

**X'(y-Xβ)=0**

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