How to proof ,

$\displaystyle

(\sin x + \cos x)\sin 2x = \sqrt 2

$

thanks

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- Dec 31st 2008, 07:47 PMSingulartrigonometry
How to proof ,

$\displaystyle

(\sin x + \cos x)\sin 2x = \sqrt 2

$

thanks - Dec 31st 2008, 08:03 PMKrizalid
You probably meant to say "how to solve" instead "how to prove," since as it's written, it just holds for certain values, in fact, it's a trig. equation.

- Dec 31st 2008, 08:05 PMLast_Singularity
$\displaystyle (\sin x + \cos x)\sin 2x = \sqrt 2$ cannot be proved because it is not true. I think that what you meant to ask is how to solve that equation for $\displaystyle x$

Try squaring both sides and expanding:

$\displaystyle (\sin x + \cos x)^2 \sin^2 2x = 2$

$\displaystyle (\sin^2 x + 2 \cos x \sin x + \cos^2 x) \sin^2 2x = 2$

$\displaystyle (1+ 2 \cos x \sin x) \sin^2 2x = 2$

$\displaystyle (1+ \sin 2x) \sin^2 2x = 2$

Letting $\displaystyle y = \sin 2x$, you have

$\displaystyle (1+ \sin 2x) \sin^2 2x = 2 \longrightarrow (1+y)y^2 = 2$

Solve for y by inspection. I see one obvious value of y such that $\displaystyle (1+y)y^2 = 2$

Then find when $\displaystyle \sin 2x$ equals that value of y and you're done.

Note: there is more than one solution. In fact, there are infinitely many solutions to this problem because sine is a periodic function. - Dec 31st 2008, 08:06 PMSingular
owh... my mistake... sorry