We have: {2x=ytan⁡θ+sin⁡θ2y=xcot⁡θ+cos⁡θ And want to prove x2+y2=1 My works: I multiplied first equation by cos⁡θ and second one by sin⁡θ and get: {2xcos⁡θ=ysin⁡θ+sin⁡θcos⁡θ2ysin⁡θ=xcos⁡θ+sin⁡θcos⁡θ By extracting sin⁡θcos⁡θ we get: 2xcos⁡θ−ysin⁡θ=2ysin⁡θ−xcos⁡θ xcos⁡θ=ysin⁡θ

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We have:  {2x=ytan⁡θ+sin⁡θ2y=xcot⁡θ+cos⁡θ  And want to prove x2+y2=1  My works:  I multiplied first equation by cos⁡θ and second one by sin⁡θ and get:  {2xcos⁡θ=ysin⁡θ+sin⁡θcos⁡θ2ysin⁡θ=xcos⁡θ+sin⁡θcos⁡θ  By extracting sin⁡θcos⁡θ we get:  2xcos⁡θ−ysin⁡θ=2ysin⁡θ−xcos⁡θ  xcos⁡θ=ysin⁡θ

Edward Patten · Accepted Answer

The system  {2xcos⁡θ=ysin⁡θ+sin⁡θcos⁡θ2ysin⁡θ=xcos⁡θ+sin⁡θcos⁡θ  is a linear system of two (independent) equations in two variables, and it is readly checked that x=sin⁡θ,   y=cos⁡θ is a solution. Therefore it is the unique solution. Now x2+y2=1 follows.

sirpsta3u · Accepted Answer

You got x=ytan⁡θ. Now substitute back into the original equations to get y=cos⁡θ  and  x=ytan⁡θ=sin⁡θ

Vasquez · Accepted Answer

We have from your last step 3xcos⁡(θ)=3ysin⁡(θ) thus x=ytan⁡(θ) putting this in (1) we get 2ytan⁡(θ)=ytan⁡(θ)+sin⁡(θ)⇒y=cos⁡(θ) from this we get x=sin⁡(θ) now note the identity that sin2⁡(θ)+cos2⁡(θ)=1

We have: {(2x=y \tan \theta+\sin \theta),(2y=x \cot \

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