Question

Verify each equation

1. cos(x + (pi/2)) = -sin x
2. sin(n pi + θ) = -1^(n) sin θ, n is an integer
3. (sin x + cos x)^(2) = 1 + sin 2x

Answer 1

1.
`\cos\left(x+\frac\pi2\right)=\cos x\cos\frac\pi2-\sin x\sin\frac\pi2=-\sin x`

2.
`\sin(n\pi+\theta)=\sin(n\pi)\cos\theta+\cos(n\pi)\sin\theta`

Since
`\{\sin0,\sin(\pm\pi),\sin(\pm2\pi),\ldots\}` all amount to 0, the first term disappears. Meanwhile,
`\{\cos0,\cos(\pm\pi),\cos(\pm2\pi),\ldots\}=\{1,-1,1,\ldots\}` in an alternating pattern, which agrees with the sequence
`(-1)^n`. Hence
`\sin(n\pi+\theta)=(-1)^n\sin\theta` for integers
`n`.

3.
`(\sin x+\cos x)^2=\sin^2x+2\sin x\cos x+\cos^2x=1+2\sin x\cos x=1+\sin2x`