(sinx)/(1+cosx)=(1-cosx)/(sinx)

simplify so RS=LS

Printable View

- Dec 12th 2011, 06:26 AMahmedbtrig identities
(sinx)/(1+cosx)=(1-cosx)/(sinx)

simplify so RS=LS - Dec 12th 2011, 06:38 AMPlatoRe: trig identities
- Dec 12th 2011, 06:39 AMahmedbRe: trig identities
thats not what i am talking about, i am talking about proofs.

like proof that RS=LS by simplifying it - Dec 12th 2011, 06:40 AManonimnystefyRe: trig identities
hi ahmedb

__Spoiler__: - Dec 12th 2011, 06:43 AMahmedbRe: trig identities
what did you multiply by 1-cosx?

y can't u use 1-sinx or anything else???

thats what confused me :S - Dec 12th 2011, 06:47 AManonimnystefyRe: 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

- Dec 12th 2011, 06:49 AMahmedbRe: trig identities
for the RHS why is it 1+cosx??

is there a method like a rule in which you have to follow? - Dec 12th 2011, 09:25 AMSironRe: trig identities
You can also use $\displaystyle \frac{a}{b}=\frac{c}{d} \Leftrightarrow a\cdot d=c\cdot b$

That means for this exercice:

$\displaystyle \frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \Leftrightarrow \sin^2(x)=[1-\cos(x)]\cdot[1+\cos(x)] \Leftrightarrow \sin^2(x)=1-\cos^2(x) \Leftrightarrow \sin^2(x)=\sin^2(x)$