How do you graph #y = sin(pi/2x + (3pi)/2)#?

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Answer 1 · 2018-01-08T15:38:54

I'm assuming you want

#y = sin(pi/2x + (3pi)/2)#

In standard form. Use the sum formula for sine, which is #sin(A + B) = sinAcosB + cosAsinB#.

#y = sin(pi/2x)cos((3pi/2)) + cos(pi/2x)sin((3pi)/2)#

As long as you know you trigonometric function values at #pi/2# and #(3pi)/2#, you're good to go.

#y = sin(pi/2x)(0) - 1(cos(pi/2x))#

#y = -cos(pi/2x)#

If you graph #f(x) = sin(pi/2x + (3pi)/2)# and #g(x) = -cos(pi/2x)#, you'll see they're indeed the same graphs.

enter image source here

Hopefully this helps!