For two different value of #theta#, #theta_1# and #theta_2#(where the value of #theta#s are between #0 and 2pi#), the equation #sin(theta+phi)=1/2sin2phi# is followed. Prove that #(sin(theta_1)+sin(theta_2)) /(costheta_1+costheta_2) =cotphi#?

Question

5637 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-07T05:16:02

Given that two different values of #theta # i.e #theta _1and theta_2# satisfy the equation

#sin(theta+phi)=1/2sin2phi#

So we have

#sin(theta_1+phi)=1/2sin2phi#

and

#sin(theta_2+phi)=1/2sin2phi#

combining these two we get

#sin(theta_1+phi)=sin(theta_2+phi)#

#=>sintheta_1cosphi+costheta_1sinphi=sintheta_2cosphi+costheta_2sinphi#

#=>sintheta_1cosphi-sintheta_2cosphi=costheta_2sinphi-costheta_1sinphi#

#=>(sintheta_1-sintheta_2)cosphi=(costheta_2-costheta_1)sinphi#

#=>(sintheta_1-sintheta_2)/(costheta_2-costheta_1)=sinphi/cosphi#

#=>(2cos((theta_1+theta_2)/2)sin((theta_1-theta_2)/2))/(2sin((theta_1-theta_2)/2)sin((theta_1+theta_2)/2))=sinphi/cosphi#

#=>cot((theta_1+theta_2)/2)=tanphi=cot((kpi)/2-phi)#,where # k=1and 3#. As #theta _1and theta_2# are between #0 and 2pi#

So

#(theta_1+theta_2)/2=((kpi)/2-phi)#

Now

#(sintheta_1+sintheta_2)/(costheta_1+costheta_2)#

#=(2sin((theta_1+theta_2)/2)cos((theta_1-theta_2)/2))/(2cos((theta_1-theta_2)/2)cos((theta_1+theta_2)/2))#

#=tan((theta_1+theta_2)/2)#

#=tan((kpi)/2-phi)=cotphi#