# Verify Trig Identities:Simplify W/Picture

• Jun 28th 2013, 07:30 PM
Melcarthus
Verify Trig Identities:Simplify W/Picture
Just simplify. See picture that is attaches. I know the even odd properties I know sin(-x) is -sinx. After that I got stuck. I tried to multiply by conjugate didnt work.
• Jun 28th 2013, 07:37 PM
Re: Verify Trig Identities:Simplify W/Picture
The answer is sin(x). So you have sin^2(-x)- sin(-x) / 1 - sin(-x) . sin^2(-x) is really just sin(-x) * sin(-x), which becomes -1*-1*sin(x)*sin(x) = sin^2 (x). So you have sin^2 (x) - sin(x) / sin(x) + 1. Take out the sin(x) on the top and you have
(sin(x)( sin(x) + 1)) / (sin(x) + 1). Cancel out the (sin(x)+1) and you have sin(x) left.
• Jun 28th 2013, 07:48 PM
Melcarthus
Re: Verify Trig Identities:Simplify W/Picture
Any advice for verifying trig identities?
• Jun 28th 2013, 07:49 PM
Melcarthus
Re: Verify Trig Identities:Simplify W/Picture
• Jun 28th 2013, 07:57 PM
Re: Verify Trig Identities:Simplify W/Picture

Does that make sense?
• Jun 29th 2013, 05:03 PM
Soroban
Re: Verify Trig Identities:Simplify W/Picture

Quote:

$\frac{\sin^2(\text{-}x) - \sin(\text{-}x)}{1-\sin(\text{-}x)}$

. . . $=\;\frac{[\sin(\text{-}x)][\sin(\text{-}x)] - (\text{-}\sin x)}{1 - (\text{-}\sin x)}$

. . . $=\;\frac{(\text{-}\sin x)(\text{-}\sin x) + \sin x}{1 + \sin x}$

. . . $=\;\frac{\sin^2\!x + \sin x}{1 + \sin x}$

. . . $=\;\frac{\sin x({\color{red}\rlap{////////}}\sin x + 1)}{{\color{red}\rlap{///////}}1 + \sin x}$

. . . $=\;\sin x$

• Jun 29th 2013, 06:55 PM
Melcarthus
Re: Verify Trig Identities:Simplify W/Picture
I got it thanks for your help both of you.