Nilai dari 2sin(π2−α)cos(π2+α)=....2\sin\left(\frac{\pi}{2}-\alpha\right)\cos\left(\frac{\pi}{2}+\alpha\right)=....2sin(2π−α)cos(2π+α)=....
2sinα−12\sin\alpha-12sinα−1
sin2α\sin2\alphasin2α
1+sin2α1+\sin2\alpha1+sin2α
1−sin2α1-\sin2\alpha1−sin2α
−sin2α-\sin2\alpha−sin2α
Persamaan di atas dapat disederhanakan dalam bentuk penjumlahan sinus
2sin(π2−α)cos(π2+α)=2sin12(π−2α)cos12(π+2α)2\sin\left(\frac{\pi}{2}-\alpha\right)\cos\left(\frac{\pi}{2}+\alpha\right)=2\sin\frac{1}{2}\left(\pi-2\alpha\right)\cos\frac{1}{2}\left(\pi+2\alpha\right)2sin(2π−α)cos(2π+α)=2sin21(π−2α)cos21(π+2α)
=2sin12(π+(−2α))cos12(π−(−2α))=2\sin\frac{1}{2}\left(\pi+\left(-2\alpha\right)\right)\cos\frac{1}{2}\left(\pi-\left(-2\alpha\right)\right)=2sin21(π+(−2α))cos21(π−(−2α))
=sinπ+sin(−2α)=\sin\pi+\sin\left(-2\alpha\right)=sinπ+sin(−2α)
Ingat kembali bahwa sin(−θ)=−sinθ\sin\left(-\theta\right)=-\sin\thetasin(−θ)=−sinθ maka
=sinπ−sin2α=\sin\pi-\sin2\alpha=sinπ−sin2α
=0−sin2α=0-\sin2\alpha=0−sin2α
=−sin2α=-\sin2\alpha=−sin2α