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

`"basically because "`

`sqrt(sin^2x+cos^2x)!=sinx+cosx`

`"note that"`

`(sinx+cosx)^2`

`=sin^2x+2sinxcosx+cos^2x`

`!=sin^2x+cos^2x`

Remember that

`y = sqrtx` if and only if `y^2 = x` (and `y >= 0`)

Why is `7 = sqrt49`? `" "` Because `7^2 = 49` (and `7 >= 0`)

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

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

It doesn't.

`(a+b)^2 = (a+b)(a+b)`

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

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

`= aa+ab+ba + b b`

`= a^2+2ab+b^2`

Which is not the same as `a^2+b^2`

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

Here's an example using numbers:

`3^2+4^2 = 9+16 = 25`

and `sqrt25 = 5` `" "` (Because `5^2 = 25`)

So `sqrt(3^2+4^2)` is not the same as `3+4`

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

`cos^2x+sin^2x-1=(cosx+sinx-1)^alpha P(sinx,cosx)`

where `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`