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C3 trig identities not in the book...

Can anyone give me a list of the identities like tan^2=sec^2 or something along those lines i know theres a few of them but i lost the paper i wrote it down on :frown:

:frown:

Original post by Vav123

Can anyone give me a list of the identities like tan^2=sec^2 or something along those lines i know theres a few of them but i lost the paper i wrote it down on :frown:

:frown:

Well tan^2(x)=sec^2(x) certainly isn't one. sec^2(x)=tan^2(x) + 1 prehaps? There's also cosec^2(x)=cot^2(x)+1 and the common sin^2(x)+cos^2(x)=1

(edited 13 years ago)

Start with

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

Dividing through by

cos^2(x)

gives

tan^2(x)+1=sec^2(x)

Dividing through by

sin^2(x)

gives

1+cot^2(x)=cosec^2(x)

tan(x)=\frac{sin(x)}{cos(x)}

Therefore

cot(x)=\frac{cos(x)}{sin(x)}

That should be it, the double angle formulae can be derived by using the addition formula provided and letting B=A

(edited 13 years ago)

Original post by dknt

Well tan^2(x)=sec^2(x) certainly isn't one. tan^(x)2=sec^2(x) + 1 prehaps? There's also cosec^2(x)=cot^2(x)+1 and the common sin^2(x)+cos^2(x)=1

That's wrong it should be

tan^2(x)+1=sec^2(x)

Original post by ElMoro

That's wrong it should be

tan^2(x)+1=sec^2(x)

My apologies!

OP

Original post by Hudzy

Start with

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

Dividing through by

cos^2(x)

gives

tan^2(x)+1=sec^2(x)

Dividing through by

sin^2(x)

gives

1+cot^2(x)=cosec^2(x)

tan(x)=\frac{sin(x)}{cos(x)}

Therefore

cot(x)=\frac{cos(x)}{sin(x)}

That should be it, the double angle formulae can be derived by using the addition formula provided and letting B=A

thanks alot :smile:

:smile:

and everyone else too

Original post by dknt

My apologies!

don't worry mate :wink2:

:wink2:

Original post by Vav123

thanks alot :smile:

:smile:

and everyone else too

Not sure which ones are needed for C3, but here are a few more simple ones:

sin(90-x)=cosx , cos(90-x)= sinx[br][br]sin(180-x)= sinx , cos(180-x)= -cosx , tan(180-x)= -tanx[br][br]sin(-x)= -sinx , cos(-x)= cosx , tan(-x)= -tanx[br][br]sin(x+y)= sinx cosy + siny cosx

and the same is true if both '+' signs are replaced with '-'.

cos(x+y)= cosx cosy - sinx siny

and again the same is true if the signs are reversed.

tan(x+y)= \frac{tanx + tany}{1 - tanx tany}

and the same holds with all signs switched.

From the above three, it is easy to derive the double angle formulae.

(edited 13 years ago)

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