Bagaimana Anda memverifikasi identitas (cosx – cosy)/(sinx + siny) + (sinx – siny)/(cosx + cosy) = 0?
Kita harus membuktikan (cosx – cosx / sinx + siny) + (sinx – siny / cosx + cosy) = 0
Larutan
Mari kita mulai dengan LHS
(cosx – nyaman / sinx + siny) + (sinx – siny / cosx + nyaman)
=[ (cosx – cosy) * (cosx + cosy) / (sinx + siny) * (cosx + cosy) + [ (sinx – siny)*(sinx + siny) / (cosx + cosy) * (sinx + siny) ] = 0
=[ (cos^2(x) + cos(x)cos(y) – cos(y)cos(x) – cos^2(y) ) / (sinx + siny) * (cosx + nyaman) + [ ( sin^2(x) + sin(x)sin(y) – sin(y)sin(x) – sin^2(y)) / (cosx + cosy) * (sinx + siny) ]
=[ (cos^2(x) + cos(x)cos(y) – cos(y)cos(x) – cos^2(y) ) + (sin^2(x) + sin(x)sin( y) – sin(y)sin(x) – sin^2(y) ) ] / [ (cosx + nyaman) * (sinx + siny) ]
=[ cos^2(x) – cos^2(y) + sin^2(x) – sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
=[ cos^2(x) + sin^2(x) – cos^2(y) – sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
Kita tahu identitasnya
cos^2(x) + sin^2(x) = 1
=[ 1 – cos^2(y) – sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
=[ 1 – (cos^2(y) + sin^2(y) ) ] / [ (cosx + nyaman) * (sinx + siny) ]
=[ 1 – (1) ] / [ (cosx + nyaman) * (sinx + siny) ]
=[ 1 – 1 ] / [ (cosx + nyaman) * (sinx + siny) ]
=[ 0 ] / [ (cosx + nyaman) * (sinx + siny) ]
0
= RHS
Maka Terbukti