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Mathematics 26 Online
OpenStudy (anonymous):

Prove frac{sin(2x)}{sqrt{2-2cos(2x)}} = sin(x)

OpenStudy (anonymous):

\[\frac{\sin(2x)}{\sqrt{2-2\cos(2x)}} = \sin(x) \]

hartnn (hartnn):

u know the formula for cos 2x ?

OpenStudy (anonymous):

and find any restrictions on x

OpenStudy (anonymous):

i actually just need that part .. the restriction

hartnn (hartnn):

ok, the denominator has square root sign and the expression under it cannot be negative. Also , denominator cannot be 0. make sense till here ?

OpenStudy (anonymous):

yes

hartnn (hartnn):

so\( 1-cos 2x \ne 0\) so \(cos2x \ne 1\) so 2x cannot equal 0,2pi,4pi,... so x cannot equal 0,pi,2pi,3pi...... ok?

OpenStudy (anonymous):

oh yeah .. cant remember properly but the test said the identity holds for 0<x<180 degrees.. is that true ?? idn if supposed to find that

hartnn (hartnn):

seems correct, because cos 2x will be negative in 3rd quadrant, then denominator can no longer be 2cos x....so yeah.

OpenStudy (anonymous):

how can i get that restriction ??

hartnn (hartnn):

u need cos 2x to be negative, right ? only then u could write denominator as 2cos x now, cos 2x negative implies, 2x is between 90 to 270 , right? so actually here i am getting x is between 45 to 135.... 45<x<135 as restriction.... does this makes sense?

OpenStudy (anonymous):

yeah i get you

hartnn (hartnn):

so overall restrictions are 45<x<135 and x \(\ne\) 0,180,360,......

OpenStudy (anonymous):

can i solve 2-2cos(2x)>0 "?

hartnn (hartnn):

if u solve that what u get?

OpenStudy (anonymous):

x>0,pi,2pi,3pi

OpenStudy (anonymous):

?

hartnn (hartnn):

oh,nopes... when we take cos 2x as negative, we already took care of expression under square root to be positive....so need of that.

OpenStudy (anonymous):

oh so you just solved 2-2cos(2x)=0 and 2-2cos(2x)<0

OpenStudy (anonymous):

so is it 0<x<180 ??

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