Question

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

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.