trig identities

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December 12th 2011, 06:26 AM
ahmedb

trig identities

(sinx)/(1+cosx)=(1-cosx)/(sinx)
simplify so RS=LS
December 12th 2011, 06:38 AM
Plato

Re: trig identities

Quote:

Originally Posted by ahmedb

(sinx)/(1+cosx)=(1-cosx)/(sinx)
simplify so RS=LS

What does $\frac{\sin(x)}{1+\cos(x)}\frac{1-\cos(x)}{1-\cos(x}}=~?$
December 12th 2011, 06:39 AM
ahmedb

Re: trig identities

thats not what i am talking about, i am talking about proofs.
like proof that RS=LS by simplifying it
December 12th 2011, 06:40 AM
anonimnystefy

Re: trig identities

hi ahmedb

Spoiler:

$ \frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \\ \\ \frac{\sin(x)\cdot(1-\cos(x))}{(1+\cos(x))(1-\cos(x))}=\frac{1-\cos(x)}{\sin(x)} \\ \\ \frac{\sin(x)\cdot(1-\cos(x))}{1-\cos^2(x)}=\frac{1-\cos(x)}{\sin(x)} \\ \\ \frac{\sin(x)\cdot(1-\cos(x))}{\sin^2(x)}=\frac{1-\cos(x)}{\sin(x)} \\ \\ \frac{1-\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)} $
December 12th 2011, 06:43 AM
ahmedb

Re: trig identities

what did you multiply by 1-cosx?
y can't u use 1-sinx or anything else???
thats what confused me :S
December 12th 2011, 06:47 AM
anonimnystefy

Re: trig identities

when you multiply by 1-sin(x) you don't get anything pretty.by multiplying by 1-cos(x) you get 1-cos^2(x) which you should now is sin^2(x).another way to do this is to multiply the fraction on the RHS by 1+cos(x). :D
December 12th 2011, 06:49 AM
ahmedb

Re: trig identities

for the RHS why is it 1+cosx??
is there a method like a rule in which you have to follow?
December 12th 2011, 09:25 AM
Siron

Re: trig identities

You can also use $\frac{a}{b}=\frac{c}{d} \Leftrightarrow a\cdot d=c\cdot b$
That means for this exercice:
$\frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \Leftrightarrow \sin^2(x)=[1-\cos(x)]\cdot[1+\cos(x)] \Leftrightarrow \sin^2(x)=1-\cos^2(x) \Leftrightarrow \sin^2(x)=\sin^2(x)$

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