which shows a correct simplification of sin(x + π)?\n○ sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·1…

which shows a correct simplification of sin(x + π)?\n○ sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·1 + cos(x)·0\n= sin(x)\n○ sin(x + π)\n= cos(x)cos(π) — sin(x)sin(π)\n= cos(x)·1 — sin(x)·0\n= cos(x)\n○ sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·-1 + cos(x)·0\n= -sin(x)\n○ sin(x + π)\n= cos(x)cos(π) — sin(x)sin(π)\n= cos(x)·-1 — sin(x)·0\n= -cos(x)

which shows a correct simplification of sin(x + π)?\n○ sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·1 + cos(x)·0\n= sin(x)\n○ sin(x + π)\n= cos(x)cos(π) — sin(x)sin(π)\n= cos(x)·1 — sin(x)·0\n= cos(x)\n○ sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·-1 + cos(x)·0\n= -sin(x)\n○ sin(x + π)\n= cos(x)cos(π) — sin(x)sin(π)\n= cos(x)·-1 — sin(x)·0\n= -cos(x)

Answer

Explanation:

Step1: Use sine addition formula

The formula for (\sin(A + B)=\sin(A)\cos(B)+\cos(A)\sin(B)). Here (A = x) and (B=\pi), so (\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)).

Step2: Substitute the values of (\cos(\pi)) and (\sin(\pi))

We know that (\cos(\pi)=- 1) and (\sin(\pi)=0). Substituting these values into the expression from Step1: (\sin(x)\times(-1)+\cos(x)\times0).

Step3: Simplify the expression

(\sin(x)\times(-1)+\cos(x)\times0=-\sin(x)+0 =-\sin(x))

Answer:

The third option: (\sin(x + \pi)) (=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)) (=\sin(x)\cdot - 1+\cos(x)\cdot0) (=-\sin(x)) is the correct simplification.