Question

Need help with Honors Pre-Calculus Questions. They have to do with the binomial theorem, DeMoivre's theorem, and a+ib form.

Answer 1

1)
ALREADY SOLVED, BUT NEED THIS ANSWER FOR THE OTHER 2

Use the binomial theorem to express (a+ib)^n in form of A+iB where A and B are summations.

Answer: \[\sum_{j=0}^{\frac{ n-1 }{ 2 }}\left(\begin{matrix}n \\ 2j\end{matrix}\right)a^{n-2j}(-1)^jb ^{2j}+i \sum_{j=0}^{\frac{ n-1 }{ 2 }}a ^{n-(2j+1)}(-1)^jb ^{2j+1} \]

Answer 2

2) Given \[t \in R, n \in N,\] use DeMoivre's theorem and the result of #1 to express:

cis(nt) = cos(nt) + isin(nt)

to where cos(nt) and sin(nt) are summations of powers of cos(t) and sin(t). Also express cos(nt) in powers of cos(t) only (using \[\cos^2 = 1-\sin^2\])

Answer 3

3) Given \[t \in R\]

Use the result of problem 2 to express, in simplest form:

A) cos(5t) in terms of powers of cos(t)
B) sin(5t) in terms of powers of sin(t)

Answer 4

Thanks a ton for the help guys. These two problems are really confusing for me and I frankly don't know where to start.

Answer 5

@ganeshie8

Answer 6

Hey, sorry to tag you but you helped me a TON last time.

Answer 7

cis(nt) = cos(nt) + isin(nt) = (cost + isint)^n

use binomial theorem ?

Answer 8

(cost + isint)^n = ?

Answer 9

You may use the result from part 1

Answer 10

Yeah that's where demoivre's theory comes in right?

Answer 11

Notice, we're using the euler formula : \(e^{it} = \cos(t) + i\sin(t)\)

Answer 12

means we can express it in terms of a power to n

Answer 13

Demoivre's theorem tells us
\((e^{it})^n = (\cos(t) + i\sin(t))^n = \cos(nt) + i\sin(nt)\)

Answer 14

Yeah

Answer 15

You're right, simply use the result from part 1 and compare the real part both sides

Answer 16

Ah. So I just plugit in to the previous formula?

Answer 17

So cos(t) would be "a" and sin(t) would be "b" and the n remains "n"

Answer 18

Yes :

\[\cos(nt) + i\sin(nt) \\~\\
= (\cos t+ i\sin t)^n\\~\\
= \sum_{j=0}^{\frac{ n-1 }{ 2 }}\left(\begin{matrix}n \\ 2j\end{matrix}\right)(\cos t)^{n-2j}(-1)^j (\sin t) ^{2j}+i \sum_{j=0}^{\frac{ n-1 }{ 2 }}(\cos t)^{n-(2j+1)}(-1)^j)(\sin t) ^{2j+1}\]

Answer 19

Compare the real parts from first line and last lines

Answer 20

Its just the +1 after the 2j, right?

Answer 21

The red parts are equal :

\[\color{red}{\cos(nt)} + i\sin(nt) \\~\\
= (\cos t+ i\sin t)^n\\~\\
= \color{Red}{\sum_{j=0}^{\frac{ n-1 }{ 2 }}\left(\begin{matrix}n \\ 2j\end{matrix}\right)(\cos t)^{n-2j}(-1)^j (\sin t) ^{2j}}+i \sum_{j=0}^{\frac{ n-1 }{ 2 }}(\cos t)^{n-(2j+1)}(-1)^j)(\sin t) ^{2j+1}\]

Answer 22

Yes, but we don't care about the imaginary part here

Answer 23

Yes that makes sense, but if we just replace one with the other then we get our initial formula

Answer 24

\[\color{Red}{\cos(nt)~~~=~~\sum_{j=0}^{\frac{ n-1 }{ 2 }}\left(\begin{matrix}n \\ 2j\end{matrix}\right)(\cos t)^{n-2j}(-1)^j (\sin t) ^{2j}}\]

Answer 25

Since you want the formula to contain only powers of cosine,
you may replace \(\color{red}{(\sin t)^{2j}}\) by \(\color{Red}{(1-\cos^2 t)^j}\) .

Answer 26

oh I see

Answer 27

Alright I got that down.

Answer 28

hello?

Answer 29

2) Is now solved, I just need help with #3

Answer 30

@ganeshie8

Answer 31

#3 is pretty straightforward after finishing #2 right ?

Answer 32

Using the result of problem 2, express:

cos5t in terms of powers of cost (simplest form)
sin5t in terms of powers of sint (simplest form)

Answer 33

Once I get a start on these problems I can normally finish them. I just never know where to start.

Answer 34

\[\color{Red}{\cos(nt)~~~=~~\sum_{j=0}^{\frac{ n-1 }{ 2 }}\left(\begin{matrix}n \\ 2j\end{matrix}\right)(\cos t)^{n-2j}(-1)^j (\sin t) ^{2j}}\]

replace n by 5

Answer 35

\[\color{Red}{\cos(5t)~~~=~~\sum_{j=0}^{\frac{ 5-1 }{ 2 }}\left(\begin{matrix}5 \\ 2j\end{matrix}\right)(\cos t)^{5-2j}(-1)^j (\sin t) ^{2j}}\]

Answer 36

\[\color{Red}{\cos(5t)~~~=~~\sum_{j=0}^{2}\left(\begin{matrix}5 \\ 2j\end{matrix}\right)(\cos t)^{5-2j}(-1)^j (\sin t) ^{2j}}\]

Answer 37

Oh, so I replace the n with the 5?

Answer 38

And now do I just expand and solve out for the 3 sums, j=0, j=1, and j=2?

Answer 39

\[\begin{align}

&\color{Red}{\cos(5t)} \\~\\
&=\color{red}{\sum_{j=0}^{2}\dbinom{5}{2j}(\cos t)^{5-2j}(-1)^j (\sin t) ^{2j}}\\~\\
&=\color{red}{ \dbinom{5}{0}(\cos t)^{5}(-1)^j (\sin t) ^{0}} + \color{red}{ \dbinom{5}{2}(\cos t)^{3}(-1)^1 (\sin t) ^{2}} + \color{red}{ \dbinom{5}{4}(\cos t)^{1}(-1)^2 (\sin t) ^{4}}\\~\\
\end{align}\]

Answer 40

Wait why did it go to \[\left(\begin{matrix}5 \\ 4\end{matrix}\right)\]

Answer 41

Oh

Answer 42

Nevermind

Answer 43

Thats 2j

Answer 44

You may simplify. Use :
\[\dbinom{n}{r} = \dfrac{n!}{(n-r)!*r!}\]

Answer 45

Also convert sin to cos using sin^2 = 1-cos^2

Answer 46

What does "r" represent?

Answer 47

oh, the bottom term.