Expert: Steve Holleran - 12/24/2007

Question
1. please write as a single trigonometric function
sin 2x/1-cos2x

2. can you determine the period and sketch the graph of

y=2sin3x

Answer
Hi Alisa,

1.  Okay, let's make use of two double-angle identities here:

sin 2x = 2 sin x cos x

cos 2x = 1 - 2sin^2 x

So,

 sin 2x / (1-cos 2x) = (2 sin x cos x) / (1 - (1-2sin^2x))

 = (2 sin x cos x) / (1 - 1 + 2sin^2 x)

 =  (2 sin x cos x) / (2 sin^2 x)

 =   cos x / sin x = cot x


2.  Well, the period is always 2 * pi / B for the sine function, where B is the coefficient of x.  So here, the period would be 2 pi / 3.

I can't sketch it here, but it would start at (0,0) and complete one wave at (2pi/3, 0).  It would cross the x-axis at pi/3; it would have a maximum of 2 at x = pi/6 and a minimum of -2 at x = pi/2.

That would be the first cycle.  If you repeat it, you would get the complete graph.

Happy holidays!

Steve