amaths

cos (x-y) = sinx siny + cosx cosy
= (2 sinx siny + 2 cosx cosy) / 2
= (2 sinx siny + 2 cosx cosy +2 -2) / 2
= (2 sinx siny + 2 cosx cosy + (sin²x + cos²x) + (sin²y + cos²y) -2) / 2
=[ (sin²x + 2sinx siny + sin²y) + (cos²x + 2cosx cosy + cos²y) – 2] / 2
= [(sin x + sin y)² + (cos x + cos y)² - 2] /2
= [(p² + q²) - 2] / 2

先將given果兩條式square both side
即係咁...
sin x+sin y =p
(sin x +sin y)^2=p^2
sin^2 x+sin^2 y+2sin x sin y=p^2---------1式

cos x+cos y =q
cos^2 x+cos^2 y+2cosy cosy=q^2----------2式

1式+2式
sin^2 x+sin^2 y+cos^2 x+sin^2 y+2sinx sin y +2cos x cos y=p^2+q^2
1+1+2(cos x cos y+sinx sin y)=p^2+q^2
2+2cos(x+y)=p^2+q^2
cos(x+y)=[(p^2 + q^2) - 2] / 2