# prove the identity

• Apr 25th 2010, 07:05 AM
slapmaxwell1
prove the identity
i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?
• Apr 25th 2010, 08:59 AM
running-gag
Hi

$\displaystyle \tan 2x - \sec 2x = \frac{\sin 2x}{\cos 2x} - \frac{1}{\cos 2x} = \frac{2 \sin x \cos x - 1}{2 \cos^2x - 1}$

$\displaystyle \tan \left(x+\frac{\pi}{4}\right) = \frac{\tan x - 1}{1 + \tan x} = \frac{\sin x - \cos x}{\cos x + \sin x}$

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

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

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

and

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

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

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

$\displaystyle \tan 2x - \sec 2x = \tan \left(x+\frac{\pi}{4}\right)$
• Apr 25th 2010, 09:42 AM
mrmohamed
Quote:

Originally Posted by slapmaxwell1
i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?

Hi all

• Apr 25th 2010, 10:31 AM
Soroban
Hello, slapmaxwell1!

This is a messy one . . . You're off to a good start!

Quote:

Prove: .$\displaystyle \tan2x - \sec2x \:=\:\tan\left(x - \tfrac{\pi}{4}\right)$

I started with the right side and used identities
. . to simplify it down to: .$\displaystyle \frac{\tan x - 1}{1 + \tan x}$

I'm not sure where to go from here? Any suggestions?

You had: .$\displaystyle \boxed{\frac{\tan x - 1}{1 + \tan x}}\;=\;\frac{\frac{\sin x}{\cos x} - 1}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\sin x - \cos x}{\cos x + \sin x} \;=\;-\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}\!:$

. . $\displaystyle -\frac{\cos x - \sin x}{\cos x + \sin x}\cdot{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \;=\;-\frac{(\cos x-\sin x)^2}{(\cos x +\sin x)(\cos x - \sin x)}$

. . . . $\displaystyle =\;-\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} - \overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\ \underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}}$

We have: . $\displaystyle -\frac{1-\sin2x}{\cos2x} \;\;=\;\;-\frac{1}{\cos2x} + \frac{\sin2x}{\cos2x}$

. . . . . . $\displaystyle =\;\;-\sec2x + \tan2x \;\;=\;\; \boxed{\tan2x - \sec2x}$

• Apr 25th 2010, 10:43 AM
Quote:

Originally Posted by slapmaxwell1
i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?

If you use the identities

$\displaystyle cos2x=\frac{1-tan^2x}{1+tan^2x}$

$\displaystyle sin2x=\frac{2tanx}{1+tan^2x}$

we obtain

$\displaystyle tan2x-sec2x=\frac{sin2x}{cos2x}-\frac{1}{cos2x}=\frac{sin2x-1}{cos2x}$

$\displaystyle =\frac{\frac{2tanx}{1+tan^2x}-1}{\left(\frac{1-tan^2x}{1+tan^2x}\right)}=\frac{\left(\frac{2tanx-1-tan^2x}{1+tan^2x}\right)}{\left(\frac{1-tan^2x}{1+tan^2x}\right)}$

$\displaystyle =\frac{tanx-1+tanx-tan^2x}{1+tan^2}\ \left(\frac{1+tan^2x}{1-tan^2x}\right)=\frac{(tanx-1)-tanx(tanx-1)}{(1-tanx)(1+tanx)}$

$\displaystyle =\frac{(tanx-1)(1-tanx)}{(1+tanx)(1-tanx)}=\frac{tanx-1}{1+tanx}$

$\displaystyle =\frac{tanx-tan\left(\frac{{\pi}}{4}\right)}{1+tan\left(\frac{ {\pi}}{4}\right)tanx}=tan\left(x-\frac{{\pi}}{4}\right)$
• Apr 25th 2010, 04:18 PM
slapmaxwell1
wow i didnt think of that approach...nice! thanks alot..so doing math i have remain open minded? LOL :O)
• Apr 25th 2010, 04:22 PM
slapmaxwell1
soroban thank you for keeping it where i could follow! whatever program your using, i want it. LOL thanks everybody for all your help..i stopped doing the problem too soon and got frustrated..and wow this was one messy problem!!!! geez

Quote:

Originally Posted by Soroban
Hello, slapmaxwell1!

This is a messy one . . . You're off to a good start!

You had: .$\displaystyle \boxed{\frac{\tan x - 1}{1 + \tan x}}\;=\;\frac{\frac{\sin x}{\cos x} - 1}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\sin x - \cos x}{\cos x + \sin x} \;=\;-\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}\!:$

. . $\displaystyle -\frac{\cos x - \sin x}{\cos x + \sin x}\cdot{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \;=\;-\frac{(\cos x-\sin x)^2}{(\cos x +\sin x)(\cos x - \sin x)}$

. . . . $\displaystyle =\;-\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} - \overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\ \underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}}$

We have: . $\displaystyle -\frac{1-\sin2x}{\cos2x} \;\;=\;\;-\frac{1}{\cos2x} + \frac{\sin2x}{\cos2x}$

. . . . . . $\displaystyle =\;\;-\sec2x + \tan2x \;\;=\;\; \boxed{\tan2x - \sec2x}$