Is $cos x + sin x - 1 = 0$ $cosx+sinx-1=0$ a root for ${cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x+sin^2x-1 = 0$ ?

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Is $cos x + sin x - 1 = 0$ $cosx+sinx-1=0$ a root for ${cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x+sin^2x-1 = 0$ ?

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Answer 1 · 2017-08-05T10:41:29

$see explanation$ $"see explanation"$

Explanation:

$basically because$ $"basically because "$

$\sqrt{{sin}^{2} x + {cos}^{2} x} \neq sin x + cos x$ $sqrt(sin^2x+cos^2x)!=sinx+cosx$

$note that$ $"note that"$

${(sin x + cos x)}^{2}$ $(sinx+cosx)^2$

$= {sin}^{2} x + 2 sin x cos x + {cos}^{2} x$ $=sin^2x+2sinxcosx+cos^2x$

$\neq {sin}^{2} x + {cos}^{2} x$ $!=sin^2x+cos^2x$

Answer 2 · 2017-08-05T16:00:40

Please see below.

Explanation:

Remember that

$y = \sqrt{x}$ $y = sqrtx$ if and only if $y^{2} = x$ $y^2 = x$ (and $y \geq 0$ $y >= 0$ )

Why is $7 = \sqrt{49}$ $7 = sqrt49$ ? $" "$ Because $7^{2} = 49$ $7^2 = 49$ (and $7 \geq 0$ $7 >= 0$ )

Many people initially think that $a + b$ $a+b$ is the same as $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ .(Until they learn that it's not.)

But that would only by true if ${(a + b)}^{2}$ $(a+b)^2$ turned out to be the same as $a^{2} + b^{2}$ $a^2+b^2$ .

It doesn't.

${(a + b)}^{2} = (a + b) (a + b)$ $(a+b)^2 = (a+b)(a+b)$

$= (a + b) (a + b)$ $= color(blue)((a+b))(color(red)(a)+color(red)(b))$

$= a (a + b) + b (a + b)$ $= color(red)(a)color(blue)((a+b))+color(red)(b)color(blue)((a+b))$

$= a a + a b + b a + b b$ $= color(red)(a)color(blue)(a)+color(red)(a)color(blue)(b)+color(red)(b)color(blue)(a)+color(red)(b)color(blue)(b)$

$= a a + a b + b a + b b$ $= aa+ab+ba + b b$

$= a^{2} + 2 a b + b^{2}$ $= a^2+2ab+b^2$

Which is not the same as $a^{2} + b^{2}$ $a^2+b^2$

(Unless one or both of $a$ $a$ and $b$ $b$ are $0$ $0$ .)

Here's an example using numbers:

$3^{2} + 4^{2} = 9 + 16 = 25$ $3^2+4^2 = 9+16 = 25$

and $\sqrt{25} = 5$ $sqrt25 = 5$ $" "$ (Because $5^{2} = 25$ $5^2 = 25$ )

So $\sqrt{3^{2} + 4^{2}}$ $sqrt(3^2+4^2)$ is not the same as $3 + 4$ $3+4$

Answer 3 · 2017-08-05T19:59:25

See below.

Explanation:

If $cos x + sin x - 1 = 0$ $cosx+sinx-1=0$ is a root of ${cos}^{2} x + {sin}^{2} x - 1$ $cos^2x+sin^2x-1$ then

${cos}^{2} x + {sin}^{2} x - 1 = {(cos x + sin x - 1)}^{α} P (sin x, cos x)$ $cos^2x+sin^2x-1=(cosx+sinx-1)^alpha P(sinx,cosx)$

where $P (sin x, cos x)$ $P(sinx,cosx)$ is a suitable non null polynomial.

Now given any $x_0 in RR, x_0 ne pi/2+2kpi$ such that $P(sinx_0,cosx_0) ne 0$ we have

$0=(cos x_0+sin x_0-1)^alphaP(sin x_0,cosx_0) rArr cos x_0+sin x_0-1=0$

because $cos^2x+sin^2x-1=0$ is an identity

but $cos x_0+sin x_0-1=0$ is not true unless $x_0 = pi/2 + 2kpi, k=0,pm1,pm2,pm3,cdots$ so it is an absurd and $cosx+sinx-1=0$ is not a root for $cos^2x+sin^2x-1$

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Is $cos x + sin x - 1 = 0$ $cosx+sinx-1=0$ a root for ${cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x+sin^2x-1 = 0$ ?

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Is $cos x + sin x - 1 = 0$ $cosx+sinx-1=0$ a root for ${cos}^{2} x + {sin}^{2} x - 1 = 0$ $cos^2x+sin^2x-1 = 0$ ?