If #(a^2-b^2)sintheta + 2ab costheta = a^2+b^2#, then what will the value of #sintheta# be?

2 Answers
Nov 27, 2016

Given

#(a^2-b^2)sintheta+2abcostheta=a^2+b^2#

#=>(a^2-b^2)/(a^2+b^2)sintheta+(2ab)/(a^2+b^2)costheta=1.....(1)#

Now if #(a^2-b^2)/(a^2+b^2)=cosalpha#

then

#sinalpha=sqrt(1-cos^2alpha)#

#=sqrt(1- ((a^2-b^2)/(a^2+b^2))^2#

#=(2ab)/(a^2+b^2)#

So the relation (1) becomes

#cosalphasintheta+sinalphacostheta=1#

#=>sin(theta+alpha)=sin(pi/2)#

#=>alpha=pi/2-theta#

#=>cosalpha=cos(pi/2-theta)#

#=>cosalpha=sintheta#

#:.sintheta=cosalpha= (a^2-b^2)/(a^2+b^2)#

Alternative-1

Given

#(a^2-b^2)sintheta+2abcostheta=a^2+b^2#

Transferring to RHS we get

#a^2(1-sintheta)+b^2(1+sintheta)-2abcostheta=0#

#=>(asqrt(1-sintheta))^2+(bsqrt(1+sintheta))^2-2ab sqrt(1-sin^2theta)=0#

#=>[(asqrt(1-sintheta))-(bsqrt(1+sintheta))]^2=0#

#=>asqrt(1-sintheta)=bsqrt(1+sintheta)#

#=>a^2(1-sintheta)=b^2(1+sintheta)#

#=>sintheta=(a^2-b^2)/(a^2+b^2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Alternative-2

Putting #b=atanphi# in the relation (1) we get

#=>(a^2-b^2)/(a^2+b^2)sintheta+(2ab)/(a^2+b^2)costheta=1#

#=>(a^2-a^2tan^phi)/(a^2+a^2tan^2phi)sintheta+(2a*atanphi)/(a^2+a^2tan^2phi)costheta=1#

#=>cos2phisintheta+sin2phicostheta=1#

#=>sin(theta+2phi)=sin(pi/2)#

#=theta=pi/2-2phi#

#=sintheta=sin(pi/2-2phi)=cos2phi#

#=>sintheta=(1-tan^2phi)/(1+tan^2phi)#

#=>sintheta=(1-(b/a)^2)/(1+(b/a)^2)=(a^2-b^2)/(a^2+b^2)#

Nov 27, 2016

#sintheta=(a^2-b^2)/(a^2+b^2)#

Explanation:

After reducing to

#(a^2-b^2)/(a^2+b^2)sintheta+(2ab)/(a^2+b^2)costheta=1#

Now

#((a^2-b^2)/(a^2+b^2))^2+((2ab)/(a^2+b^2))^2=(a^4+2a^2b^2+b^4)/(a^4+2a^2b^2+b^4)=1# so there exists #phi# such that

#sinphi=(a^2-b^2)/(a^2+b^2)# and
#cosphi=(2ab)/(a^2+b^2)#

note that

#-1 le (a^2-b^2)/(a^2+b^2) le 1#
#-1 le (2ab)/(a^2+b^2) le 1#

so

#sinthetasinphi+costhetacosphi=1#

and #theta = phi# because #sin^2theta+cos^2theta=1# then

#sintheta=sinphi=(a^2-b^2)/(a^2+b^2)#