Решите эту тригонометрические уравнения

[paha]

CrypticMath в сообщении #393951 писал(а):

Should mathematical induction be used?

This is not the case!

CrypticMath в сообщении #393951 писал(а):

Why is my initial approach incorrect?

Becuase $\log$ is not homomorphism:

$$ \log(a-b)\not=\log{a}-\log{b} $$

-- Пт дек 31, 2010 02:09:15 --

CrypticMath
You can extract a solution from the above discussion (using only international mathematical notations)

[CrypticMath]

(Оффтоп)

Эта проблема, кажеться не из области регулярной математики старшей средней школы.Первоначально казалось что это простые трегонометрические уравнения

Is this all that's required?
For $n > 2$, the biggest that $\cos^n - \sin^n$ can be is:
$|\cos^n (x)| \leq \cos^2 (x), |\sin^n (x)| \leq \sin^2 (x)$

$|\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \cos^2 (x) + \sin^2 (x) = 1$

[paha]

CrypticMath в сообщении #394053 писал(а):

s this all that's required?
For $n > 2$, the biggest that $\cos^n - \sin^n$ can be is:
$|\cos^n (x)| \leq \cos^2 (x), |\sin^n (x)| \leq \sin^2 (x)$

$|\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \cos^2 (x) + \sin^2 (x) = 1$

yes... $n\ge 2$

And then the equality $|\cos^n (x) - \sin^n (x)|=1$ holds?

The case $n=1$ must be analyzed apartly

[ewert]

paha в сообщении #394115 писал(а):

$n\ge 2$

$n>2$ (при $n=2$ логика немножко другая). И вообще неравенства в цепочке должны быть строгими, иначе логика не работает.

[CrypticMath]

paha писал(а):

The case $n=1$ must be analyzed apartly

For n = 1, use the addition formula:
$C\ \sin(x + \theta) = C\left[ \sin x\ \cos \theta + \cos x\ \sin \theta\right] = C\ \sin \theta\ \cos x + C \cos \theta\ \sin x$

$A = C\ \sin \theta,\ B = C \cos \theta.$
$\mathrm{When}\ A = 1, B = -1$:
$\displaystyle \cos x - \sin x = \sqrt{2}\ \sin\left(x + \frac{3\pi}{4}\right)$

$\mathrm{Then}\ \cos x - \sin x = 1 \Leftrightarrow$

$\displaystyle \sin \left(x + \frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}},\,\, x + \frac{3\pi}{4} = \frac{3\pi}{4}\ \mathrm{or}\ \frac{\pi}{4},\,\, x = - \frac\pi2\,\,\mathrm{or}\ 0$

[CrypticMath]

Is the problem finished?

[paha]

You solve the case $n=1$.

So, if $n\ge 2$ then

CrypticMath в сообщении #394053 писал(а):

$|\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \cos^2 (x) + \sin^2 (x) = 1$

And?

[CrypticMath]

paha в сообщении #394372 писал(а):

You solve the case $n=1$.

So, if $n\ge 2$ then

CrypticMath в сообщении #394053 писал(а):

$|\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \cos^2 (x) + \sin^2 (x) = 1$

And?

$ = |\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \sout{\cos^2 (x) + \sin^2 (x)} = 1$

$ = |\cos^n(x) - \sin^n(x)| \leq |\cos^n(x)|+|\sin^n(x)|\leq 1$

[paha]

CrypticMath в сообщении #394402 писал(а):

$ = |\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \sout{\cos^2 (x) + \sin^2 (x)} = 1$

$ = |\cos^n(x) - \sin^n(x)| \leq |\cos^n(x)|+|\sin^n(x)|\leq 1$

And?

Where are the roots?

-- Вс янв 02, 2011 13:09:51 --

like this ($n=1$)

CrypticMath в сообщении #394233 писал(а):

$ x + \frac{3\pi}{4} = \frac{3\pi}{4}\ \mathrm{or}\ \frac{\pi}{4},\,\, x = - \frac\pi2\,\,\mathrm{or}\ 0$

[CrypticMath]

Sonic86 писал(а):

Случай $n=2k$ простой.
В случае $n=2k+1$ можно сделать замену $t=-x$ и потом попытаться использовать основное тригонометрическое тождество.

paha писал(а):

$|\cos^n (x) - \sin^n (x)| \leq |\cos^n (x)| + |\sin^n (x)| \leq \cos^2 (x) + \sin^2 (x) = 1$
And?

OK, here's another attempt:

$\cos^{2k} x = 1 + \sin^{2k} x.$

Roots: $x = 0$ or $\pi$

For $n = 2k + 1.\ \mathrm{Set}\ t = -x$:

$1 = \cos^{2k + 1} x - \sin^{2k + 1} x = \cos^{2k + 1} (-t) - \sin^{2k + 1} (-t)$

$= \cos^{2k + 1} t + \sin^{2k + 1} t$

$\leq |\cos^{2k + 1} y| + |\sin^{2k + 1} y|$

$= \cos^2 t|\cos^{2k - 1} t| + \sin^2|\sin^{2k + 1} t|$

$\leq \cos^2 t + \sin^2 t = 1.$

$\displaystyle \mathrm{Roots:}\ \cos t = 1\ \mathrm{and}\ x = 2k\pi\ \mathrm{or}\ \sin t = 1\ \mathrm{and}\ x = 2k\pi - \frac\pi2$

[CrypticMath]

Check my last post please?
...finished?

[paha]

CrypticMath в сообщении #394653 писал(а):

$\cos^{2k} x = 1 + \sin^{2k} x$
but $\cos^{2k}\le 1$ so $\sin x=0$, therefore
Roots: $x = 0$ or $\pi$

CrypticMath в сообщении #394653 писал(а):

For $n = 2k + 1.\ \mathrm{Set}\ t = -x$:

$1 = \cos^{2k + 1} x - \sin^{2k + 1} x = \cos^{2k + 1} (-t) - \sin^{2k + 1} (-t)$

$= \cos^{2k + 1} t + \sin^{2k + 1} t$

$\leq |\cos^{2k + 1} y| + |\sin^{2k + 1} y|$

$= \cos^2 t|\cos^{2k - 1} t| + \sin^2|\sin^{2k + 1} t|$

$\leq \cos^2 t + \sin^2 t = 1.$

So all inequalities in this chain are equalities. But $ \cos^2 t|\cos^{2k - 1} t| + \sin^2t|\sin^{2k + 1} t|=\cos^2 t + \sin^2 t$ is possible iff either $\cos{t}=0$ or $\sin{t}=0$, therefore...

[CrypticMath]

paha писал(а):

So all inequalities in this chain are equalities. But $ \cos^2 t|\cos^{2k - 1} t| + \sin^2t|\sin^{2k + 1} t|=\cos^2 t + \sin^2 t$ is possible iff either $\cos{t}=0$ or $\sin{t}=0$, therefore...

I'm not sure?!

...the solutions are: $\displaystyle x = 2k\pi\ and\ 2k\pi - \frac{\pi}{2}$

[paha]

CrypticMath в сообщении #394958 писал(а):

...the solutions are: $\displaystyle x = 2k\pi\ and\ 2k\pi - \frac{\pi}{2}$

do not harry!

If $\cos{t}=0$ then $\cos x=0$ then $\sin{x}=\pm 1$ then...

-- Пн янв 03, 2011 21:08:03 --

And the same with $\sin{t}=0$

[CrypticMath]

paha в сообщении #394966 писал(а):

CrypticMath в сообщении #394958 писал(а):

...the solutions are: $\displaystyle x = 2k\pi\ and\ 2k\pi - \frac{\pi}{2}$

do not harry!

If $\cos{t}=0$ then $\cos x=0$ then $\sin{x}=\pm 1$ then...

-- Пн янв 03, 2011 21:08:03 --

And the same with $\sin{t}=0$

I think: $\displaystyle \cos t = 0\ \mathrm{in\ the\ interval}\ \frac\pi2$

$\displaystyle \sin x = \pm1\ \mathrm{in\ the\ intervals}\ \frac\pi2, -\frac\pi2$