Cho $cosx+cosy=p;sinx+siny=q(p ²+q ² khác 0)$.Tính $sin(x+y)$ HELP ME, PLEASE :(

Đáp án:

 `sin(x+y)={2pq}/{p^2+q^2}`

Giải thích các bước giải:

 `\qquad cosx+cosy=p`

`=>cos^2x+2cosxcosy+cos^2y=p^2`

`\qquad sinx+siny=q`

`=>sin^2x+2sinxsiny+sin^2y=q^2`

`=>(sin^2x+cos^2x)+(sin^2y+cos^y)+2(cosxcosy+sinxsiny)=p^2+q^2`

`=>1+1+2cos(x-y)=p^2+q^2`

`=>cos(x-y)={p^2+q^2-2}/2`

$\\$

Ta có:

`\qquad (sinx+siny).(cosx+cosy)=qp`

`<=>sinxcosx+sinycosy+(sinxcosy+sinycosx)=pq`

`<=>1/ 2 sin2x+1/ 2 sin2y+sin(x+y)=pq`

`<=>sin(x+y)+1/ 2 (sin2x+sin2y)=pq`

`<=>sin(x+y)+1/ 2 . 2.sin\ {2x+2y}/2 . cos\ {2x-2y}/2=pq`

`<=>sin(x+y)+sin(x+y).cos(x-y)=pq`

`<=>sin(x+y).[1+cos(x-y)]=pq`

`<=>sin(x+y)={pq}/{1+cos(x-y)}`

`<=>sin(x+y)`$=\dfrac{pq}{1+\dfrac{p^2+q^2-2}{2}}$

`<=>sin(x+y)={2pq}/{p^2+q^2}`

Vậy `sin(x+y)={2pq}/{p^2+q^2}`