椭圆x^2/16+y^2/4=1有两点P、Q.O是原点,若OP、OQ斜率乘积为-1/4.证明:|OP|^2+|OQ|^2

椭圆x^2/16+y^2/4=1有两点P、Q.O是原点,若OP、OQ斜率乘积为-1/4.证明:|OP|^2+|OQ|^2为定值.
设P(4cosp,2sinp),Q(4cosq,2sinq),
OP、OQ斜率乘积为sinpsinq/(4cospcosq)=-1/4,
∴cospcosq+sinpsinq=0,
∴cos(p-q)=0,
∴|OP|^2+|OQ|^2=12[(cosp)^2+(cosq)^2]+8
=6[2+cos2p+cos2q]+8
=6[2+2cos(p+q)cos(p-q)]+8
=20.
请问这一步是怎么推出来的?
=6[2+cos2p+cos2q]+8
=6[2+2cos(p+q)cos(p-q)]+8