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## Topic review (newest first)

NIH
2005-06-18 07:35:16

In a non-right angled triangle, with internal angles a, b, c, tan a + tan b + tan c = tan a * tan b * tan c.

MathsIsFun
2005-06-14 08:38:53

sin (x/2) = ± sqrt[ (1 - cos x) / 2 ]

cos (x/2) = ± sqrt[ (1 + cos x) / 2 ]

(This is exactly what mathsyperson had, just with a square root applied)

The "±" means that you need to provide your own + or - depending on where the angle points.

... and there are lots more "trigonometric identities" ... too many for me to remember ... I had to look this one up.

mathsyperson
2005-06-14 06:14:53

Not sure if this is what you wanted but...

sin^2(x/2) = (1-cos x)/2
cos^2(x/2) = (1+cos x)/2

I think to get half angles like that you can substitute in cos x and then square root what you get.

Doggy
2005-06-14 04:50:48

pople please help
cos x/2=?
sin x/2=?

btw
sin(x+/-y)=sin^x ^ cos^y +/- cos^x ^ sin^y

Mr T
2005-05-03 06:32:34

i knew that one.

Roraborealis
2005-04-30 22:06:51

When do you learn this?

MathsIsFun
2005-04-30 08:56:29

Hey, mathsyperson, please register and swing by every so often, you could help solve some of the visitor's problems with that knowledge.

Here is a trick to remember:  Sohcahtoa  ... sounds like an American Indian tribe, but can help you out at exams:

Sine = Opposite / Hypotenuse         (Soh...)
Cosine = Adjacent / Hypotenuse      (...cah...)
Tangent = Opposite / Adjacent         (...toa)

mathsyperson
2005-04-30 05:58:25

cos = adjacent divided by hypotenuse
tan = opposite divided by adjacent
tan = sin divided by cos
sec = 1 divided by cos
cosec = 1 divided by sin
cot = 1 divided by tan OR cos divided by sin

umm...

1 + cot^2x = cosec^2x
tan^2x + 1 = sec^2x

Roraborealis
2005-04-30 00:52:54

Sin= opposite divided by hypotenuse!

That's all I know.....

Zach
2005-04-29 05:51:30

Yes, I did think. But, it's not a thought allowable for this forum.

mathsyperson
2005-04-29 02:00:28

Ooh, ooh, I know one!

sin^2x + cos^2x = 1

So there!

einstein
2005-04-29 00:02:43

ha ha ha ha!  u thought!!

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