Trig Derivation Help??!

I have no idea. Who says you can't? Do it anyway. And good luck. I hate math.

haha me too. but it's a question in my math homework and i need an answer. i derived it another way, using this division thingie, but i just need a smart sounding explanation for this one, even if it's bs.

any help would be great...

I've read this but....I can't help.

You, too, McT? lol

This is a good thread.

I like "Confunction Identity".

It's a nice phrase.

They won't have heard of it in Lydford though.

I believe it's cofunction, not confunction.

The cofunction of tan(a) is cot(90-a), st tan(a)=cot(90-a)

This can be shown from the definition of tan and cot, and the cofunctions of sin(a) and cos(a)

remember that sin(a)=cos(90-a) & cos(a)=sin(90-a) (cofunction of sin and cos)

tan(a)=sin(a)/cos(a)=cos(90-a)/sin(90-a)=cot(90-a)

similarly you can show that

cot(a)=tan(90-a)

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Wow what lvl maths is this? university?...it seems pretty complicated.

I wouldn't think it would be college level. Probably second year of high school algebra, or trigonometry when discussing relationships between trig functions.

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what! when i did igcse (11th grade international cambridge examination) there was nothing about cofunction identities...man you americans must be smart..

It's just a name, that quickly falls out of the relation between sin and cos when you apply the pythagorean theorem (sin^2x+cos^2x+1) and the multiple angle relations. Tan an cot is the same thing.

Although the word may be new, the relation is probably known. The only extrapolation I did here was to recognized that confunction must be cofunction.

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it

The question is ill posed.

/Kayyam