JohnLight
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#1
Report Thread starter 10 years ago
#1
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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Concept186
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#2
Report 10 years ago
#2
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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viksta1000
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#3
Report 10 years ago
#3
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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disco1000
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#4
Report 10 years ago
#4
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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ella37
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#5
Report 10 years ago
#5
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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JohnLight
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#6
Report Thread starter 10 years ago
#6
(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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JohnLight
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#7
Report Thread starter 10 years ago
#7
Thanks everyone who replied, i'm sorry you had to witness my momentary ******edness.
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immense010
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#8
Report 10 years ago
#8
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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ZeeAli
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#9
Report 10 years ago
#9
depends i think what x is
cosX +sinX = 1
where X=0
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ddrrzzeerr
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#10
Report 10 years ago
#10
If that were the case then why would anyone bother quoting the squared version?
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marek35
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#11
Report 10 years ago
#11
Generally- no.

Also the square root of cos^2x is not equal to cosx.
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amy2507
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#12
Report 10 years ago
#12
No
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cabdirisaq
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#13
Report 2 years ago
#13
Sec-CosX÷1 SinX
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cabdirisaq
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#14
Report 2 years ago
#14
SecX-CosX÷1+sinx
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gdunne42
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#15
Report 2 years ago
#15
(Original post by cabdirisaq)
SecX-CosX÷1+sinx
Why have you resurrected an 8 year old discussion to make an unrelated comment?
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Younosewho
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#16
Report 1 year ago
#16
Yo been 8 years do u even remember visiting this site?What do u do now?
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