unit test\nunit test review active\nwhich shows a correct simplification of sin(x + π)?\nsin(x + π)\n=…

unit test\nunit test review active\nwhich shows a correct simplification of sin(x + π)?\nsin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x) · 1 + cos(x) · 0\n= sin(x)\nsin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x) · 1 - sin(x) · 0\n= cos(x)\nsin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x) · -1 + cos(x) · 0\n= -sin(x)\nsin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x) · -1 - sin(x) · 0\n= -cos(x)

unit test\nunit test review active\nwhich shows a correct simplification of sin(x + π)?\nsin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x) · 1 + cos(x) · 0\n= sin(x)\nsin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x) · 1 - sin(x) · 0\n= cos(x)\nsin(x + π)\n= sin(x)cos(π) + cos(x)sin(π)\n= sin(x) · -1 + cos(x) · 0\n= -sin(x)\nsin(x + π)\n= cos(x)cos(π) - sin(x)sin(π)\n= cos(x) · -1 - sin(x) · 0\n= -cos(x)

Answer

Explanation:

Step1: Recall the sine addition formula

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

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

We know that (\cos(\pi)=- 1) and (\sin(\pi)=0). Substitute into (\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)) [ \begin{align*} \sin(x+\pi)&=\sin(x)\times(-1)+\cos(x)\times0\ &=-\sin(x)+0\ &=-\sin(x) \end{align*} ]

Answer:

The correct simplification is (\sin(x+\pi)=\sin(x)\cos(\pi)+\cos(x)\sin(\pi)=-\sin(x)) (assuming the option that follows this calculation is correct among the given choices in the original multiple - choice setup).