ATAR Notes: Forum
HSC Stuff => HSC Subjects + Help => Topic started by: Lydia.k on June 24, 2019, 07:45:21 am
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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
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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.)
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Thanks a lot :) ;D