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Math Help - when does cos^2x=sin^2x?

  1. #1
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    when does cos^2x=sin^2x?

    when does cos^2 x=sin^2 x?
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  2. #2
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    Quote Originally Posted by Amberosia32 View Post
    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.
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    Quote Originally Posted by Amberosia32 View Post
    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....
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    Hello, Amberosia32!

    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

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  5. #5
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    cos^2 x = sin^2 x?
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