嗯,如此看來,LZ應(yīng)該不知道二倍角公式和和差化積、積化和差,先介紹介紹.
二倍角公式：cos(2x)=(cosx)^2-(sinx)^2=2(cosx)^2-1=1-2(sinx)^2
和差化積：cos(x+y)+cos(x-y)=2cosxcosy,即
cosp+cosq=2cos[(p+q)/2]cos[(p-q)/2]
積化和差就是逆用和差化積!
①3-4sin^2 (60-y)
=1+2[1-2(sin60-y)^2]=1+2cos(120-2y)……注：LZ在這里應(yīng)該是多打了個平方!
=2[1/2+cos(120-2y)]=2[cos60+cos(120-2y)]……因為cos60=0.5……
=4cos(90-y)cos(y-30)…………和差化積
=4sinycos(30-y)……普通三角恒等變換
②cos^2(30-b)+cos^2(30-y)-2cos(30-b) cos(30-y)cos a
=[1+cos(60-2b)]/2+[1+cos(60-2y)]/2-[cos(60-y-b)+cos(y-b)]cosa逆用二倍角公式,用積化和差
=1+[cos(60-2b)+cos(60-2y)]/2-cos(60-y-b)cosa-cos(b-y)cosa
=1+[cos(60-2b)+cos(60-2y)]/2-(cosa)^2-cos(b-y)cosa……這里是不是有已知條件y=c?才使得
cos(60-y-b)=cos(60-c-b)=cosa?
=1+cos(60-b-y)cos(b-y)-(cosa)^2-cos(b-y)cosa…………和差化積
=1+cosacos(b-y)-(cosa)^2-cos(b-y)cosa……這里是不是有已知條件y=c?才使得
cos(60-y-b)=cos(60-c-b)=cosa?
=1-(cosa)^2……第二項和第四項抵消
回答完畢!