when does cos^2x=sin^2x?

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February 4th 2010, 02:26 PM
Amberosia32

when does cos^2x=sin^2x?

when does cos^2 x=sin^2 x?
February 4th 2010, 02:32 PM
Prove It

Quote:

Originally Posted by Amberosia32

when does cos^2 x=sin^2 x?

You should know from the Pythagorean Identity that

$\cos^2{x} + \sin^2{x} = 1$ .

So $\cos^2{x} = 1 - \sin^2{x}$ .

Substituting this into your original equation:

$\cos^2{x} = \sin^2{x}$

$1 - \sin^2{x} = \sin^2{x}$

$1 = 2\sin^2{x}$

$\sin^2{x} = \frac{1}{2}$

$\sin{x} = \pm\frac{1}{\sqrt{2}}$

$x = \left \{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right \} + 2\pi n$ , where $n$ is an integer representing the number of times you have gone around the unit circle.
February 4th 2010, 02:48 PM
Archie Meade

Quote:

Originally Posted by Amberosia32

when does cos^2 x=sin^2 x?

$Cos^2x=Sin^2x$

$Cos^2x-Sin^2x=0$

$\left(Cosx+Sinx\right)\left(Cosx-Sinx\right)=0$

$Cosx=Sinx,\ or\ Cosx=-Sinx$

As Cosx gives the horizontal co-ordinate and Sinx gives the vertical co-ordinate of the unit circle centred at the origin,

Cosx=Sinx at $\frac{\pi}{4}+2n\pi,\ \left(\pi+\frac{\pi}{4}+2n\pi\right)$ for n=0,1,2.....

Cosx= -Sinx at $\left(\pi-\frac{\pi}{4}+2n\pi\right),\ \left(2{\pi}-\frac{\pi}{4}+2n\pi\right)$ for n=0.1.2....
February 4th 2010, 05:43 PM
Soroban

Hello, Amberosia32!

Quote:

When does $\cos^2\!x \:=\:\sin^2\!x\;?$

We have: . $\sin^2\!x \:=\:\cos^2\!x$

Divide by $\cos^2\!x\!:\;\;\;\frac{\sin^2\!x}{\cos^2\!x} \:=\:1 \quad\Rightarrow\quad \left(\frac{\sin x}{\cos x}\right)^2 \:=\:1 \quad\Rightarrow\quad \tan^2\!x \:=\:1$ . $\Rightarrow\quad \tan x \:=\:\pm1$

Therefore: . $x \;=\;\frac{\pi}{4} + \frac{\pi}{2}n\;\;\text{ for any integer }n$
February 4th 2010, 10:22 PM
pacman

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cos^2 x = sin^2 x?