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

Aug 5, 2017

$\text{see explanation}$

Explanation:

$\text{basically because }$

$\sqrt{{\sin}^{2} x + {\cos}^{2} x} \ne \sin x + \cos x$

$\text{note that}$

${\left(\sin x + \cos x\right)}^{2}$

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

$\ne {\sin}^{2} x + {\cos}^{2} x$

Aug 5, 2017

Explanation:

Remember that

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

Why is $7 = \sqrt{49}$? $\text{ }$ Because ${7}^{2} = 49$ (and $7 \ge 0$)

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 ${\left(a + b\right)}^{2}$ turned out to be the same as ${a}^{2} + {b}^{2}$.

It doesn't.

${\left(a + b\right)}^{2} = \left(a + b\right) \left(a + b\right)$

$= \textcolor{b l u e}{\left(a + b\right)} \left(\textcolor{red}{a} + \textcolor{red}{b}\right)$

$= \textcolor{red}{a} \textcolor{b l u e}{\left(a + b\right)} + \textcolor{red}{b} \textcolor{b l u e}{\left(a + b\right)}$

$= \textcolor{red}{a} \textcolor{b l u e}{a} + \textcolor{red}{a} \textcolor{b l u e}{b} + \textcolor{red}{b} \textcolor{b l u e}{a} + \textcolor{red}{b} \textcolor{b l u e}{b}$

$= a a + a b + b a + b b$

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

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 $\sqrt{25} = 5$ $\text{ }$ (Because ${5}^{2} = 25$)

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

Aug 5, 2017

See below.

Explanation:

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

${\cos}^{2} x + {\sin}^{2} x - 1 = {\left(\cos x + \sin x - 1\right)}^{\alpha} P \left(\sin x , \cos x\right)$

where $P \left(\sin x , \cos x\right)$ is a suitable non null polynomial.

Now given any ${x}_{0} \in \mathbb{R} , {x}_{0} \ne \frac{\pi}{2} + 2 k \pi$ such that $P \left(\sin {x}_{0} , \cos {x}_{0}\right) \ne 0$ we have

$0 = {\left(\cos {x}_{0} + \sin {x}_{0} - 1\right)}^{\alpha} P \left(\sin {x}_{0} , \cos {x}_{0}\right) \Rightarrow \cos {x}_{0} + \sin {x}_{0} - 1 = 0$

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

but $\cos {x}_{0} + \sin {x}_{0} - 1 = 0$ is not true unless ${x}_{0} = \frac{\pi}{2} + 2 k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \cdots$ so it is an absurd and $\cos x + \sin x - 1 = 0$ is not a root for ${\cos}^{2} x + {\sin}^{2} x - 1$