Question

sin x - cos x = 1/2

x aparține (0; pi/2)

Calculati:

sin 2x
cos 2x
tg (x/2)​

Answer 1

Răspuns:

$x$∈$(0,\frac{\pi }{2})=> cosx > 0, sin x>0, tgx>0$

$a) sinx-cosx=\frac{1}{2}$ (Ridicam la patrat)

$sin^2x+cos^2x-2sinx*cosx=\frac{1}{4}$

$1-2sinx*cosx=\frac{1}{4} =>-2sinx*cosx=-\frac{3}{4}$

$2sinx*cosx=\frac{3}{4}$

$2sinx*cosx=sin(2x)=\frac{3}{4}$

$b) cos2x=cos^2x-sin^2x=(cosx-sinx)(cosx+sinx)=-\frac{1}{2} *(cosx+sinx)$

Facem separat cosx+sinx:

$cosx+sinx=\sqrt{(cosx+sinx)^2}=\sqrt{cos^2x+sin^2x+2sinx*cosx}\\=\sqrt{1+\frac{3}{4} }=\sqrt{\frac{4}{4}+\frac{3}{4} }=\sqrt{\frac{7}{4} } =\frac{\sqrt{7} }{2}$

Inlocuim:

$cos2x=-\frac{1}{2}*\frac{\sqrt{7} }{2}=-\frac{\sqrt{7} }{4}$

$c) tg(\frac{x}{2})$

Stim ca $tg(\frac{x}{2})=\frac{1+sinx-cosx}{1+sinx+cosx}$

Facem separat 1+sinx-cosx si 1+sinx+cosx

$1+sinx-cosx=1+\frac{1}{2} =\frac{3}{2}$

$1+\frac{\sqrt{7} }{2} =\frac{2+\sqrt{7} }{2}$

Inlocuim:

$\frac{\frac{3}{2} }{\frac{2+\sqrt{7} }{2} } =\frac{3}{2+\sqrt{7} }$(Amplificam cu conjugata lui 2+$\sqrt{7}$, adica 2-$\sqrt{7}$)

$\frac{3(2-\sqrt{7}) }{(2+\sqrt{7})(2-\sqrt{7}) } =\frac{6-3\sqrt{7} }{4-7} =\frac{6-3\sqrt{7} }{-3} =-\frac{3(2-\sqrt{7}) }{3} =-2+\sqrt{7}$

$tg(\frac{x}{2} )=-2+\sqrt{7}$

Obs.: Pentru a putea rezolva si ultimul ex. a trebuit sa folosesc formula

$\frac{1+sinx-cosx}{1+sinx+cosx}$, atlfel nu se putea, dar nu te speria ptr. ca e destul de usor de demonstrat...Te vei obisnui