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}`
