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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!!