Hey,

I am not sure how to sketch the following, as was wondering if anyone could help me?

Sketch y=sin(x+pi) for 0<X<2pi

Cheers,

Lydia

Start by sketching \(y = f(x)\), where \(f(x) = \sin x\).

Then \(f(x+\pi) = \sin(x+\pi)\), so you’re after the curve \(y=f(x+\pi)\). Recall that this is a

**translation** of \(y=f(x)\) to the left by \(\pi\) units, so that’s what huh need to sketch.

(Note that the graph will correspond to that of \(y=-\sin x\). This is not a coincidence, because \(\sin(x+\pi)=-\sin x\) is a trigonometric identity for third quadrant angles.)