Trig functino help - tiny question

K.ourt

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Jul 20, 2005
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17
kay, so I have this equation: y=2sin3(x+(pi/6))-1 that I have to take through a table of transformations.

I can get from y=sin(x) to y=2sin(x), but from there I cant get to y+2sin3(x).

It would be a horizontal stretch by a factor of (1/3), right? Meaning I would multiply the x values by (1/3).

So for (pi/2) x (1/3) would I get like 6pi? But that doesnt check..
 
K.ourt said:
kay, so I have this equation: y=2sin3(x+(pi/6))-1 that I have to take through a table of transformations.

I can get from y=sin(x) to y=2sin(x), but from there I cant get to y+2sin3(x).

It would be a horizontal stretch by a factor of (1/3), right? Meaning I would multiply the x values by (1/3).

So for (pi/2) x (1/3) would I get like 6pi? But that doesnt check..


For cosine the period will change to: 2π3\displaystyle \frac{2 \pi}{3}

Period change is found by (Old period)/(what is multiplying the angle)
 
Hello, K.ourt!

Let's take baby-steps . . .

I have this equation: y=2sin[3(x+π6)]1\displaystyle y\:=\:2\cdot\sin\left[3\left(x\,+\,\frac{\pi}{6}\right)\right]\,-\,1

that I have to take through a table of transformations.
We already know the graph of: y=sinx\displaystyle \,y\:=\:\sin x

Now, we start "inside" the function.

The +π6\displaystyle \,+\,\frac{\pi}{6} moves the graph π6\displaystyle \frac{\pi}{6} units to the left.

The 3\displaystyle \,3 compresses the cycle by a factor of 13\displaystyle \frac{1}{3}.

The 2\displaystyle \,2 gives it an amplitude of 2.

And the 1\displaystyle \,-1 lowers the graph one unit.
 
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