acos²;θ+asin²;θ=? 是acos²;θ+bsin²;θ=?

acos²;θ+asin²;θ=? 是acos²;θ+bsin²;θ=?


acos²;θ+asin²;θ
=a(cos²;θ+sin²;θ)
=a



求證4sinαcosα(cos²;α-sin²;α)=sin4α左=


左=4sinαcosα(cos²;α-sin²;α)=2sin2a·cos2a=sin4α