Quick Trig Prob Question

[us3rnotfound]

Say I have the equation y = sin2(x + pi/3).

is the period equal to pi, or is it equal to 2pi?

Thanks

 

[us3rnotfound]

Also, in this case, is 2 the amplitude?

 

[RaynorWolfcastle]

sin2 = sin^2?
If so:
sin^2(x+pi/3) = 0.5*{1+cos[2(x+pi/3)]}
hence, periodic with period pi.

 

[us3rnotfound]

Originally posted by: RaynorWolfcastle
sin2 = sin^2?

no

 

[us3rnotfound]

ttt

 

[Toastedlightly]

sounds like you need a double angle formula.... or something. I dunno!

 

[esun]

The period of y = sin(2*x + 2*pi/3) is the same as the period for y = sin(2*x). The only difference between these is a phase shift of 2*pi/3. The amplitude is one, as denoted here: y = 1*sin(2*x).

So the period of y = sin(2*x) is given by some basic math. We need to find when 2*x = 2*pi, which is when x = pi, hence the period is pi. This makes sense because if you think logically, at x = 0, y = sin(2*0) = 0. At x = pi, y = sin(2*pi) = 0. And, obviously, at half the period it is also zero.

Edit for a typo.

 

[TuxDave]

Originally posted by: us3rnotfound
Also, in this case, is 2 the amplitude?

No

 

[RaynorWolfcastle]

Originally posted by: us3rnotfound

Originally posted by: RaynorWolfcastle
sin2 = sin^2?

no

your notation is unclear then...

If by y = sin2(x + pi/3). you actually mean sin[2(x+pi/3)], the period is pi
If by y = sin2(x + pi/3). you actually mean [sin(x+pi/3)]^2, the period is pi (as I showed above)
If by y = sin2(x + pi/3). you actually mean 2*sin(x+pi/3), the period is 2pi

I think that covers all the contingencies.