Answer :
1. [tex]\cos\left(x+\dfrac\pi2\right)=\cos x\cos\dfrac\pi2-\sin x\sin\dfrac\pi2=-\sin x[/tex]
2. [tex]\sin(n\pi+\theta)=\sin(n\pi)\cos\theta+\cos(n\pi)\sin\theta[/tex]
Since [tex]\{\sin0,\sin(\pm\pi),\sin(\pm2\pi),\ldots\}[/tex] all amount to 0, the first term disappears. Meanwhile, [tex]\{\cos0,\cos(\pm\pi),\cos(\pm2\pi),\ldots\}=\{1,-1,1,\ldots\}[/tex] in an alternating pattern, which agrees with the sequence [tex](-1)^n[/tex]. Hence [tex]\sin(n\pi+\theta)=(-1)^n\sin\theta[/tex] for integers [tex]n[/tex].
3. [tex](\sin x+\cos x)^2=\sin^2x+2\sin x\cos x+\cos^2x=1+2\sin x\cos x=1+\sin2x[/tex]
2. [tex]\sin(n\pi+\theta)=\sin(n\pi)\cos\theta+\cos(n\pi)\sin\theta[/tex]
Since [tex]\{\sin0,\sin(\pm\pi),\sin(\pm2\pi),\ldots\}[/tex] all amount to 0, the first term disappears. Meanwhile, [tex]\{\cos0,\cos(\pm\pi),\cos(\pm2\pi),\ldots\}=\{1,-1,1,\ldots\}[/tex] in an alternating pattern, which agrees with the sequence [tex](-1)^n[/tex]. Hence [tex]\sin(n\pi+\theta)=(-1)^n\sin\theta[/tex] for integers [tex]n[/tex].
3. [tex](\sin x+\cos x)^2=\sin^2x+2\sin x\cos x+\cos^2x=1+2\sin x\cos x=1+\sin2x[/tex]