# Identities and solve for x

• Nov 1st 2007, 06:14 PM
Freaky-Person
Identities and solve for x
Man our whole class struggled at this one the whole period.

Solve for x

1. $\displaystyle tanx + sec2x = 1$

Prove

2. $\displaystyle (\frac{\sin4x}{1-cos4x})*(\frac{1-cos 2x}{cos2x}) = tanx$

3. $\displaystyle cot^2A * sin^2 (\frac{\pi}{2}-A) = tan(\frac{\pi}{2}- A) - cosA$
• Nov 1st 2007, 07:04 PM
Freaky-Person
OOOhh! nevermind question 1!

I just remembered that after 2 hours of deliberation between everyone in the class + 1, I finally got it.

I got up to $\displaystyle \frac{(cos2x + 1)(sin2x)}{cos^2 2x} = tan x$ on question 2

And $\displaystyle \frac{\frac{cos^3 A}{sinA}}{1+sinA} = \frac{sinA + sin^2 A - 1}{cosA}$

On the last one.

• Nov 1st 2007, 07:05 PM
Soroban
Hello, Freaky-Person!

I assume you know the Double- and Half-angle Identities.
. . Here's #2 . . .

Quote:

2) Prove: .$\displaystyle \frac{\sin 4x}{1-\cos4x}\cdot \frac{1-\cos 2x}{\cos2x} \;= \;\tan x$

We have: .$\displaystyle \frac{2\sin2x\cos2x}{2\sin^2\!2x}\cdot\frac{2\sin^ 2\!x}{\cos2x} \;=\;\frac{2\sin^2\!x}{\sin2x} \;= \;\frac{2\sin^2\!x}{2\sin x\cos x} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x$

• Nov 1st 2007, 10:16 PM
DivideBy0
Quote:

Originally Posted by Freaky-Person
3. $\displaystyle cot^2A * sin^2 (\frac{\pi}{2}-A) = tan(\frac{\pi}{2}- A) - cosA$

My calculator says this is false.
• Nov 1st 2007, 10:39 PM
kalagota
Quote:

Originally Posted by DivideBy0
My calculator says this is false.

yeah, i think..
i was able to show
$\displaystyle cot^2A - cos^2A \neq cot A - cos A$
• Nov 2nd 2007, 06:23 AM
Soroban
Hello, Freaky-Person!

Quote:

1. Solve for $\displaystyle x\!:\;\;\tan x + \sec2x \:= \:1$

We have: .$\displaystyle \frac{\sin x}{\cos x} + \frac{1}{\cos2x} \;=\;1$

. . $\displaystyle \sin x\cos2x + \cos x \;=\;\cos x\cos2x$

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

. . $\displaystyle \sin x - 2\sin^3\!x + \cos x \;=\;\cos x - 2\sin^2\!x\cos x$

. . $\displaystyle 2\sin^3\!x - 2\sin^2\!x\cos x - \sin x \;=\;0$

. . $\displaystyle \sin x\left(2\sin^2\!x - 2\sin x\cos x - 1\right) \;=\;0$

And we have two equations to solve . . .

$\displaystyle [1]\;\;\sin x \:=\:0\quad\Rightarrow\quad\boxed{ x \:=\:0,\,\pm\pi,\,\pm2\pi, \cdots}$

$\displaystyle [2]\;\;2\sin^2x - 2\sin x\cos x - 1 \;=\;0$

Divide by $\displaystyle \cos^2\!x\!:\;\;2\tan^2\!x - 2\tan x - \sec^2\!x \;=\;0$

. . $\displaystyle 2\tan^2\!x - 2\tan x - (\tan^2\!x + 1) \;=\;0$

. . $\displaystyle \tan^2\!x - 2\tan x - 1 \;=\;0$

Quadratic Formula: .$\displaystyle \tan x \;=\;\frac{2 \pm\sqrt{2^2-4(1)(-1)}}{2(1)} \;=\;1 \pm \sqrt{2}$

. . Hence: .$\displaystyle x \;=\;\arctan\left(1 \pm\sqrt{2}\right)$

Therefore: .$\displaystyle \boxed{x \:=\:\begin{Bmatrix}\frac{3\pi}{8} + 2\pi n \\ \text{-}\frac{\pi}{8} + 2\pi n \end{Bmatrix}}$