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    If cos^2 x + sin^2 x = 1, does cos x + sin x = 1? I'm not sure because, cos^2 x = (cosx)^2 therefore when you take the square root you get cos x. So if you take the square root of everything in the trig identity cos^2 x + sin^2 x = 1 you get cos x + sin x = 1. I dont think this is right but i dont know what i'm doing wrong.
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    The square root of cos^2 + sin^2 isn't cos + sin. Think about this: 3^2 + 4^2 = 5^2 (9+16=25). Does 3 + 4 = 5?
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    nope doesnt work

    Sin45 = 1/r2

    co45 = 1/r2

    1/r2 + 1/r2 =/= 1

    this is because the square root of sin^2x+cos^2x =/= sinx+cosx
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    Your fundamentally abusing the laws of powers in this case.
    When you sqrt (sin^2+cos^2 = 1) you square root one and get +/- 1 and when you square root the other side, because its + and not times or divide, you can't just split up the two terms. Immagine that (cos^2 (x) and sin^2 (x)) are a bracket, you can't just square root both sides.

    Immagine if you could. Then you are saying that sin (x) + cos (x) = sqrt (sin^2 (x) + cos^2 (x)) so if we square both sides, we'd get the same value on either side right? then you get sin^2 (x) + cos^2(x) + 2sin(x)cos(x) = sin^2 (x) + cos^2(x) thus 2sin(x)cos(x) = 0. so it only works if sin (x) or cos (x) are 0 and for no other value.
    Good enough
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    if you're taking the square root, you take the WHOLE equation. you can't just square root cosx^2 and sinx^2 separately. that's only when the variable are multiplied, not added.
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    (Original post by Concept186)
    The square root of cos^2 + sin^2 isn't cos + sin. Think about this: 3^2 + 4^2 = 5^2 (9+16=25). Does 3 + 4 = 5?
    Thanks man, i feel like such an idiot!
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    Thanks everyone who replied, i'm sorry you had to witness my momentary retardedness.
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    Although it's not true generally, there are values of x where it does work, those being in any multiple of pi/2. Here's how it works:
    cos(x)+sin(x)=1

(cos(x)+sin(x))^2=1^2=1

cos^2(x)+sin^2(x)+2cos(x)sin(x)=  1

cos^2(x)+sin^2(x)=1-2cos(x)sin(x).
    So what you're saying only works when:
    2cos(x)sin(x)=0

cos(x)=0 or sin(x)=0
    x=\frac{\pi}{2} ,\frac{3\pi}{2} ,etc...
    OR
    x=0, \pi, 2\pi, etc..
    i.e. x=\frac{n \pi}{2} where n=0,1,2,...
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    depends i think what x is
    cosX +sinX = 1
    where X=0
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    If that were the case then why would anyone bother quoting the squared version?
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    Generally- no.

    Also the square root of cos^2x is not equal to cosx.
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    No
 
 
 
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