Expert: Scotto - 6/28/2008

Question
QUESTION: Good day Scotto, thanks for helpin out on these 2 qns that have been bugging me lately...

1)How do we graph r=1-2sin(3theta) and r=1+2sin(theta/2)?

I understand we need knowledge of polar coords and complex numbers, which are my weak areas. Hence, I would appreciate much if you explain your line of thoughts in solving these too...

2) use deMoivre's Theorem to express sin(5theta) in terms of sin(theta)..



thanks lots!

ANSWER: 1) First, know what the sin curve looks like.  It is zero at every npi for all integers n.  It curves up in the first interval and down in the second.  From the third on, it repeats this pattern.

The 3theta in there would make this up and down motion go 3 times as fast (it would go up and down 3 times on 0 to 2pi instead of just once).  In the y direction, sin goes from -1 to 1, so 2sin goes from
-2 to 2, and then 1-2sin goes -1 to 3.

Dividing theta by 2 would make the interval of repetition twice as long.  Multiplying by 2 would have the same effect.  Adding to 1 instead of subtracting from 1 would make it look like a mirror image.  Instead of going up in the first interval, it would go down.

2)deMoivre's theorem states that

 (cosx + isinx)^n = cos(nx) + isin(nx).

I don't see how that could be used to express sin(5theta) in terms of sin(theta) alone, for what I get is

 sin(5x) = ((cosx + isinx)^n - cos(nx))/i.

You can find more information on deMoivre's Theorem  in http://en.wikipedia.org/wiki/De_Moivre's_formula .


---------- FOLLOW-UP ----------

QUESTION: thanks scotto for the help...i do have follow-up on both answers tho...

1) i understand how the curves would look like, but I have a solution sheet that suggests that for r=1-2sin(theta), the sketch of this eqn is a pattern around the origin...the same is true for r=1+2sin(theta/2).

Thus, is there any other way of drawing the curves to the eqns?

2) The solution manual suggests

(cos(theta)+isin(theta))^5=cos(5theta) + isin(5theta)
=cos^5(theta)+i5cos^4(theta)sin(theta)-10cos^3(theta)sin^2(theta)
-i10cos^2(theta)sin^3(theta)+5cos(theta)sin^4(theta)+isin^5(theta)

Comparing the imaginary parts,
sin5(theta)=5cos^4(theta)sin(theta)-10cos^2(theta)sin^3(theta)+sin^5(theta)

= 5(1-sin^2(theta))^2- 10(1-sin^2(theta))sin^3(theta)+ sin^5(theta)
= 5sin(theta)-20sin^3(theta)+ 16sin^5(theta)

I havent been able to understand this solution, could you help make sense of it?

thanks lots for the help :)

Answer
1) Note that both of the curves start at (0,1) since sin(0) is 0.  Doing this problem we will take Equation 1 as r=1-2sin(theta) and Equation 2 as r=1+2sin(theta/2).

Equation 1 would start by going down to -1 and then up to 3.  The second curve would go in the opposite direction.  Equation 1 would go up and down twice as fast as equation 2.

To draw them make note of the maximums, minimums, and interval of repetition.  Not that for Equation 2 the interval takes twice as long as equation 1.

2) Express a number in complex mathematics as a number with a magnitude and an angle.  When a power n is taken, the magnitude is taken to the n and the angle is multiplied by n.

For example, let's look at the 2nd, 3rd, and 6th roots of -64.  The magnitude is 64 and the andgle is 180 (0 degrees goes along positive real axis).  For the powers, we can take the magnitude to that power.

The squareroot root is 8, the cube root is 4, and the sixth root is 2.
This can be used to determine the magnitude.

For the squareroot, the magnitude was 2 the angles are 180/2=90 and (180+360)/2=270 (straight down).

For the cube root, the angle is 180/3=60, (180+360)/3 = 180, and (180+2*360)/3=300.

For the sixth root, the angles would be be (180+360n)/6 where n is 0,1,2,3,4 and 5.

It is nice to know the number's found in Pascal's triangle in computing the sum of two variables to a power, but not needed here.
You know, 1, 1 1, 1 2 1, 1 3 3 1, 1 4 6 4 1, 1 5 10 10 5 1, 1 6 15 20 15 6 1, etc.

Best wishes!