Question

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

Answer 1

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)`

Answer 2

`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)`