# Trig issues...

• May 5th 2006, 09:59 PM
Pyotr
Trig issues...
I am completely and utterly stuck on this one problem:

http://img72.imageshack.us/img72/252...846541acb8.png

I am supposed to verify this identity using only the fundamental identities. Could somebody at least lead me in the right way to start this? I have no clue where to even begin with this one :(
• May 5th 2006, 10:15 PM
CaptainBlack
Quote:

Originally Posted by Pyotr
I am completely and utterly stuck on this one problem:

http://img72.imageshack.us/img72/252...846541acb8.png

I am supposed to verify this identity using only the fundamental identities. Could somebody at least lead me in the right way to start this? I have no clue where to even begin with this one :(

Here we go:

$
\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x) }=\frac{(\sin(x))^2+(1+cos(x))^2}{(1+\cos(x))\sin( x)}
$

$
=\frac{(\sin(x))^2+1+2\cos(x)+(cos(x))^2}{(1+\cos( x))\sin(x)}=$
$\frac{\{(\sin(x))^2+(cos(x))^2\}+1+2\cos(x)}{(1+\c os(x))\sin(x)}
$

$
=\frac{2+2\cos(x)}{(1+\cos(x))\sin(x)}=\frac{2}{ \sin (x)}
$
• May 26th 2006, 09:05 PM
chancey
I did it this way:

$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x) }=\frac{2}{\sin(x)}$

$\frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}$

$\sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0$

$(\sin(x) - (1+\cos(x)))^2 = 0$

$(\sin(x) - \cos(x) - 1)^2 = 0$
• May 27th 2006, 12:32 AM
CaptainBlack
Quote:

Originally Posted by chancey
I did it this way:

$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x) }=\frac{2}{\sin(x)}$

$\frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}$

$\sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0$

$(\sin(x) - (1+\cos(x)))^2 = 0$

$(\sin(x) - \cos(x) - 1)^2 = 0$

RonL
• May 27th 2006, 12:36 AM
CaptainBlack
Quote:

Originally Posted by chancey
I did it this way:

$\frac{\sin(x)}{1+\cos(x)}+\frac{1+\cos(x)}{\sin(x) }=\frac{2}{\sin(x)}$

$\frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1 + \cos(x))} = \frac{2}{\sin(x)}$

$\sin^2(x)+(1+\cos(x))^2 - 2\sin(x)(1+\cos(x)) = 0$

$(\sin(x) - (1+\cos(x)))^2 = 0$

$(\sin(x) - \cos(x) - 1)^2 = 0$

Even if this were correct it would not as it stands constitute a proof that
the first line is an identity, because if this were correctly done you will
have shown that if you assume the first line is an identity then the last
line is an identity (which if you had done this right it of course would have
been).

You will need to work the other way around. Start with a known identity
and from that derive the equation you wish to show is an identity.

RonL
• May 27th 2006, 12:55 AM
chancey
Woops, thanks for pointing that out, I didn't to multiply the other side of the fraction, I thought the answer was strange

$\frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1+\cos(x))} = \frac{2}{\sin(x)}$

$\sin(x)(\sin^2(x)+(1+\cos(x))^2) - 2\sin(x)(1+\cos(x)) = 0$

$\sin^2(x)+(1+\cos(x))^2 - 2(1+\cos(x)) = 0$

$\sin^2(x)+1+2\cos(x)+\cos^2(x) - 2-2\cos(x) = 0$

$1+1+2\cos(x) - 2-2\cos(x) = 0$

$0 = 0$
• May 27th 2006, 01:26 AM
CaptainBlack
Quote:

Originally Posted by chancey
Woops, thanks for pointing that out, I didn't to multiply the other side of the fraction, I thought the answer was strange

$\frac{\sin^2(x)+(1+\cos(x))^2}{\sin(x)(1+\cos(x))} = \frac{2}{\sin(x)}$

$\sin(x)(\sin^2(x)+(1+\cos(x))^2) - 2\sin(x)(1+\cos(x)) = 0$

$\sin^2(x)+(1+\cos(x))^2 - 2(1+\cos(x)) = 0$

$\sin^2(x)+1+2\cos(x)+\cos^2(x) - 2-2\cos(x) = 0$

$1+1+2\cos(x) - 2-2\cos(x) = 0$

$0 = 0$

My other comment still applies.

RonL
• May 27th 2006, 01:35 AM
chancey
Quote:

Originally Posted by CaptainBlack
My other comment still applies.

I don't understand what you mean? The steps show that the equation is correct and I did use one of the trig identities to simplify...

EDIT: Ohhh I just understood what you meant as I clicked the button, you mean I need to manipulate the LHS to equal the RHS to show that they are equal?
• May 27th 2006, 01:40 AM
CaptainBlack
Quote:

Originally Posted by chancey
I don't understand what you mean? The steps show that the equation is correct and I did use one of the trig identities to simplify...

EDIT: Ohhh I just understood what you meant as I clicked the button, you mean I need to manipulate the LHS to equal the RHS to show that they are equal?

Exactly.

RonL