which shows a correct simplification of sin(x + π)?\no 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)?\no sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·1 + cos(x)·0\n= sin(x)\no sin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x)·1 - sin(x)·0\n= cos(x)\no sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)· -1 + cos(x)·0\n= -sin(x)\no 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)?\no sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)·1 + cos(x)·0\n= sin(x)\no sin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x)·1 - sin(x)·0\n= cos(x)\no sin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x)· -1 + cos(x)·0\n= -sin(x)\no 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: Apply sum - formula for sine

The sum - formula for sine is $\sin(A + B)=\sin(A)\cos(B)+\cos(A)\sin(B)$. Here $A = x$ and $B=\pi$, so $\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)$.

Step2: Evaluate $\cos(\pi)$ and $\sin(\pi)$

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

Answer:

The correct simplification is $\sin(x + \pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)=\sin(x)\times(-1)+\cos(x)\times0=-\sin(x)$, so the correct option is the one that shows $\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)=\sin(x)\cdot(-1)+\cos(x)\cdot0 =-\sin(x)$.