# Establish the Identity

• Aug 24th 2008, 02:56 PM
magentarita
Establish the Identity
Establish the identity.

[sin^3 (x) + cos^3 (x)]/[1 - 2 cos^2 (x)] =
[sec(x) - sin(x)]/[tan(x) - 1]
• Aug 24th 2008, 03:32 PM
Jhevon
Quote:

Originally Posted by magentarita
Establish the identity.

[sin^3 (x) + cos^3 (x)]/[1 - 2 cos^2 (x)] =
[sec(x) - sin(x)]/[tan(x) - 1]

Consider the LHS:

$\displaystyle \frac {\sin^3 x + \cos^3 x}{1 - 2 \cos^2 x} = \frac {(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{- \cos 2x}$

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

Now consider the RHS:

$\displaystyle \frac {\sec x - \sin x}{\tan x - 1} = \frac {\frac 1{\cos x} - \sin x}{\frac {\sin x}{\cos x} - 1}$

$\displaystyle = \frac {\frac {1 - \sin x \cos x}{\cos x}}{ \frac {\sin x - \cos x}{\cos x}}$

$\displaystyle = \frac {1 - \sin x \cos x}{\sin x - \cos x}$

$\displaystyle = \frac {1 - \sin x \cos x}{\sin x - \cos x} \cdot \frac {\sin x + \cos x}{\sin x + \cos x}$

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

$\displaystyle = LHS$

Thus, $\displaystyle LHS \equiv RHS$
• Aug 24th 2008, 03:43 PM
skeeter
actually, this is not an identity ... the equation is not true for odd multiples of
$\displaystyle x = \frac{\pi}{2}$
• Aug 24th 2008, 04:00 PM
Soroban
Hello, magentarita!

Quote:

Establish the identity: .$\displaystyle \frac{\sin^3\!x + \cos^3\!x}{1 - 2\cos^2\!x}\;= \;\frac{\sec x - \sin x}{\tan x - 1}$
The left side has a sum of cubes ... factor it.

. . . $\displaystyle \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{1 - 2 \cos^2\!x}$

. . $\displaystyle = \;\frac{(\sin x + \cos x)(\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}} - \sin x\cos x)}{\underbrace{(1 - \cos^2\!x)}_{\text{This is }\sin^2\!x} - \cos^2\!x}$

. . $\displaystyle = \;\frac{(\sin x + \cos x)(1 - \sin x\cos x)}{\sin^2\!x - \cos^2\!x}$

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

. . $\displaystyle = \;\frac{1 - \sin x\cos x}{\sin x - \cos x}$

Divide top and bottom by $\displaystyle \cos x$

. . $\displaystyle \frac{\dfrac{1}{\cos x} - \dfrac{\sin x{\color{red}\rlap{/////}}\cos x}{{\color{red}\rlap{/////}}\cos x}} {\dfrac{\sin x}{\cos x} - \dfrac{{\color{red}\rlap{/////}}\cos x}{{\color{red}\rlap{/////}}\cos x}}$

. . $\displaystyle = \;\frac{\sec x - \sin x}{\tan x - 1}$

• Aug 25th 2008, 02:53 AM
magentarita
Thanks...
I thank you all, especially Jhevon and Soroban.
• Aug 25th 2008, 07:21 AM
Jhevon
Quote:

Originally Posted by magentarita
I thank you all, especially Jhevon and Soroban.

skeeter had a valid point as well. the secant is undefined for the specified values of x. so that may cause problems. after graphing the two functions, i saw that the graphs were the same and so i overlooked it. i guess the point here was to do the algebraic manipulation to show you know your trig identities and how to use them
• Aug 25th 2008, 08:54 PM
magentarita
No doubt...
Quote:

Originally Posted by Jhevon
skeeter had a valid point as well. the secant is undefined for the specified values of x. so that may cause problems. after graphing the two functions, i saw that the graphs were the same and so i overlooked it. i guess the point here was to do the algebraic manipulation to show you know your trig identities and how to use them

No doubt but to a struggling math student, the more detail given the better chance for understanding.