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ha ha ha ha! u thought!!
Ooh, ooh, I know one!
sin^2x + cos^2x = 1
So there!
Yes, I did think. But, it's not a thought allowable for this forum.
Boy let me tell you what:
I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.
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Sin= opposite divided by hypotenuse!
That's all I know.....
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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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
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)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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When do you learn this?
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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i knew that one.
I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend
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pople please help
cos x/2=?
sin x/2=?
btw
sin(x+/-y)=sin^x ^ cos^y +/- cos^x ^ sin^y
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.
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.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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In a non-right angled triangle, with internal angles a, b, c, tan a + tan b + tan c = tan a * tan b * tan c.
2 + 2 = 5, for large values of 2.
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