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Precalculus
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Which function's graph has a period of 2? y=cos(x-pi/2) y=3cos pi x y=2 sin 3x y=-4 sin 2x
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The function takes the form \[\huge y=a\sin(bx+c)+d\]in general. By the way, it doesn't matter here, if sin is replaced by cos. The period is given by this formula \[\huge \frac{2\pi}{b}\]
so it would be y= 2 sin 3x
no it's not y = 2*sin(3x)
that has a period of T = 2pi/B T = 2pi/3
y=3cos pi x? 2 pi/pi
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Yes, that's right ^.^
\[\Large \frac{2\pi}{\pi}=2\]
thank you for explaining how to do it
y= -4 sin pi x
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