(sin 3x + sin x) / (cos 3x + cos x) = sqrt 3, sin x = ?
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"If the momentum of an object increases by #20%#, what will be the percent increase in its kinetic energy?"
#sin x=1/2# [when #0<=x<=pi/2#]
#(sin 3x + sin x) / (cos 3x + cos x) = sqrt 3#
We'll apply 2 identities here, #color(red)(sin3x = 3sinx-4sin^3x)# and #color(red)(cos3x=4cos^3x-3cosx)#
or, #(color(red)(3sinx-4sin^3x)+sinx)/(color(red)(4cos^3x-3cosx)+cosx)=sqrt3#
Combining #sinx# terms,
or, #(4sinx-4sin^3x)/(4cos^3x-2cosx)= sqrt3#
or, #[4sinx(1-sin^2x)]/[2cosx(2cos^2x-1)]=sqrt3#
Again, 2 more identities, #color(magenta)(cos^2x = 1-sin^2x# and #color(magenta)(cos2x=2cos^2x-1#
or, #(2sinxcancel(color(magenta)(cos^2x))^(cosx))/(cancelcosxcolor(magenta)(cos2x))= sqrt3#
or, #(2sinxcosx)/(cos2x)=sqrt3#
One more here, #color(blue)(sin2x=2sinxcosx#
or, #color(blue)(sin2x)/(cos2x)= sqrt3#
or, #tan2x= sqrt3#
or, #tan2x= tan(pi/3)#
or,# 2x= pi/3#
or, #x= pi/6#
so, #sinx= sin(pi/6)= 1/2#
We may use,
#rarrsinA+sinB=2sin((A+B)/2)*cos((A+B)/2)# and
#rarrcosA+cosB=2cos((A+B)/2)*cos((A+B)/2)#
Given that,
#rarr(sin3x+sinx)/(cos3x+cos)=sqrt(3)#
#rarr(cancel(2)sin((3x+x)/2)*cancel(cos((3x-x)/2)))/(cancel(2)cos((3x+x)/2)*cancel(cos((3x-x)/2)))=sqrt(3)#
#rarrtan2x=tan60°#
#rarr2x=60°#
#rarrx=30°#
Now, #sinx=sin30°=1/2#
