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**einstein****Guest**

ha ha ha ha! u thought!!

**mathsyperson****Guest**

Ooh, ooh, I know one!

sin^2x + cos^2x = 1

So there!

**Zach****Member**- Registered: 2005-03-23
- Posts: 2,075

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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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

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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**mathsyperson****Guest**

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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)

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

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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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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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**Doggy****Guest**

pople please help

cos x/2=?

sin x/2=?

btw

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

**mathsyperson****Guest**

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.

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

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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