Maths help please! Clarify #14 Thanks!

[Deleted]

Why does 

12 (cos 5pi/3 + i sin 5pi/3) = 12 ( cos pi/3 - i sin pi/3) 

Thats the next step in my solution manual, but I can't figure out why? 

[yogiii]

I think because you have to look at exact values. e.g. cos pi/3 is 1/2 and cos 5pi/3 is also 1/2

[Green Eyed Lady]

Why? Because....world peace.

[MarleyKeezy78]

I hope this helps  

[chiver78]

Feb 23, 2016 13:09:41 GMT -5 yogiii said:

I think because you have to look at exact values. e.g. cos pi/3 is 1/2 and cos 5pi/3 is also 1/2

and on the flip side, your sin terms are .866... and -.866....

I started to look at this earlier and got pulled away. where I had gotten stuck was trying to figure out what i had to do with anything. maybe that's where you're getting stuck?

[Deleted]

Feb 23, 2016 13:09:41 GMT -5 yogiii said:

I think because you have to look at exact values. e.g. cos pi/3 is 1/2 and cos 5pi/3 is also 1/2

This makes sense. They are equal. I'm not sure why I would know to convert them to those equalities though if I hadn't had the book to show me? I will go back and look at it now with this in mind though. Thanks! 

[yogiii]

That's part of math, problem solve in any way you can!

[Virgil Showlion]

Remember that e^(theta i) = cos (theta) + i sin(theta)

Hence:

12 (cos 5pi/3 + i sin 5pi/3) = 12 e^(5/3 pi i)

Also remember that e^(2 pi i) = 1 and thus e^(-2 pi i) = 1

Thus 12 e^(5/3 pi i) 
    = 12 e^(5/3 pi i) · 1
    = 12 e^(5/3 pi i) · e^(-2 pi i)
    = 12 e^(5/3 pi i) · e^(-6/3 pi i)
    = 12 e^(5/3 pi i - 6/3 pi i)
    = 12 e^(-1/3 pi i)
    = 12 (cos (-1/3 pi) + i sin(-1/3 pi))

Finally, remember that cos (-theta) = cos(theta) and sin (-theta) = -sin(theta)

Hence,

12 (cos (-1/3 pi) + i sin(-1/3 pi)) = 12 (cos pi/3 - i sin pi/3)

[chiver78]

the post I started earlier had a question for you - would you recognize that those terms were in radians? there are 2pi radians in 360° - so 5pi/3 = 5/3 pi => 300°

read through this, it's a pretty well written explanation.
Radians

[Peace Of Mind]

I was hoping to learn something but all that happened was this:

[GRG a/k/a goldenrulegirl]

Show-offs.  

[cael]

[Miss Tequila]

<<slowly backs out of the nerd thread>>

[cael]

I may have a STEM degree but the 'M' I was able to satisfy with "calculus for life sciences" which I was lucky to get a C in.

[Deleted]

Ok. Let me try to ask this better.

The problem was Multiply (6 cis 4pi/3) (2 cis pi/3) and express the answer in rectangular form. Give an exact answer.

He got

(6 cis 4pi/3) (2 cis pi/3) = 12 cis 5 pi/3 = 12 (cos 5pi/3 + i sin 5pi/3)

Now, if I was just going to figure this out to a decimal point I could just put it in the calculator. BUT. The way they want the answer is to use a Sin=Opposite/Hypotenuse etc type thing on a 60/30/90 triangle, so since 5pi/3 = 300 degrees, he can't use that. He needs to convert it to pi/3 to get a 60 degree so he could use square root of 3/2 for sin.

Now. I know that 12 (cos 5pi/3 + i sin 5pi/3) = 12 ( cos pi/3 - i sin pi/3) because I have the solutions manual.

But what steps should he be following to get
from .... 12 (cos 5pi/3 + i sin 5pi/3) to ...... 12 ( cos pi/3 - i sin pi/3)

How does one know they substitute? Is there a table I should be using?

Or is this something that just really doesn't matter eventually since most actual applications of this type of problem will use decimals and not be limited to expressions of known angles in a Sin = Opposite/Hypotenuse etc type manner?

Does that make sense? It really has been a long day...

[Virgil Showlion]

oped : If the steps in Reply #7 don't make sense, all that's needed is a few trigonometric identities. In particular:

1. cos( -theta ) = cos( theta )
2. sin( -theta ) = -sin( theta )
3. cos( theta + 2 n pi ) = cos( theta ), where n is any integer
4. sin( theta + 2 n pi ) = sin( theta ), where n is any integer

We start with: 

12 (cos 5pi/3 + i sin 5pi/3)

Expanding this and reordering terms (this is just my preference):

12 cos(5/3 pi) + 12 i sin(5/3 pi)

Now, identities 3 and 4 tell us that adding any integer multiple of 2 pi to the argument of cos() or sin() won't change the value of the functions. (2 pi radians is 360°, exactly one full revolution, as chiver points out. If we do any integer number of full revolutions, our "directional arrow" representing the hypotenuse is pointing in the same direction as it was before. This is the intuition behind identities 3 and 4.). -1 is an integer, hence we can add -2 pi to the arguments of the cos() and sin() functions without changing their values.

In doing so, we get:

12 cos(5/3 pi - 2 pi) + 12 i sin(5/3 pi - 2 pi)

Now we simplify. First, by converting the fractions to like denominators:

12 cos(5/3 pi - 6/3 pi) + 12 i sin(5/3 pi - 6/3 pi)

Then by performing the subtraction:

12 cos(-1/3 pi) + 12 i sin(-1/3 pi)

Now we take advantage of trigonometric identities 1 and 2 to compute cos(-1/3 pi) and sin(-1/3 pi). In particular:

cos(-1/3 pi) = cos(1/3 pi)
sin(-1/3 pi) = -sin(1/3 pi)

Substituting these in, we get

12 cos(1/3 pi) - 12 i sin(1/3 pi)

And factoring out the 12 gives us

12(cos(1/3 pi) - i sin(1/3 pi))

or as you've written it:

12 ( cos pi/3 - i sin pi/3)

This technique will work regardless of the specific angles you're using. That's what makes trigonometric identities so useful. You don't need to compute numerical values for cos() and sin() at any point.

ETA: FWIW, the numerical values in this particular case are:

cos( pi/3 ) = cos( 60° ) = 1/2
sin( pi/3 ) = sin( 60° ) = sqrt(3)/2 ~= 0.866

[justme]

Oh dear, it's been so long since I've done anything with cos and sin (except ya know, like staying up all night drinking and cursing kind). I guess that's all on the engineering side of math? Definitely not using it in finance!

[svwashout]

I think you can get there in two steps. First remember that the cycle length of sin and cos is 2pi,
so you can add or subtract any multiple of 2pi to the argument of either sin or cos and get the same result.

So if we subtract 2pi from both arguments:
   12(cos(5pi/3) + i sin(5pi/3))
= 12(cos(-pi/3) + i sin(-pi/3))

The next step is to remember that the sign of sin follows its argument but the sign of cos doesn't
care about its argument, so cos(-Z) = cos(Z) but sin(-Z) = -sin(Z). This means if we flip the
-pi/3 to +pi/3, the cos term doesn't change sign but the sin term flips to negative:

= 12(cos(+pi/3) - i sin(+pi/3)).