# Thread: complex numbers and trigonometry

1. ## complex numbers and trigonometry

Given that $(1+ itan@)^n + (1-itan@)^n = 2sec^n@cosn@$
prove that the real part of $(1+itan\frac {\pi}{8})^8 = 64(12\sqrt2 - 17)$

thanks

2. Originally Posted by differentiate
Given that $(1+ itan@)^n + (1-itan@)^n = 2sec^n@cosn@$
prove that the real part of $(1+itan\frac {\pi}{8})^8 = 64(12\sqrt2 - 17)$

thanks
Exactly what do you know? More precisely do you know that:

$\text{re}( (1+i \tan(\theta))^n )= (1/2)[(1+ i\tan(\theta))^n + (1-i\tan(\theta))^n]$?

This is because if we write $z=re^{i\phi}$ then:

$z^n=r^n e^{i n \phi}$

so:

$\text{re} z^n = r^n \cos(n \phi)$

but:

$z^n+\overline{z}^n= r^n[e^{i n \phi}+e^{-i n \phi}]=2 r^n \cos(n \phi)$.

CB

3. A more fun way

By Bernoulli expansion,

(1-itanA))^8 = (1)^8 + 8(1)^7(-itanA) + 28(1)^6(-itanA)^2 + 56(1)^5(-itanA)^3 + 70(1)^4(-itanA)^4 + 56(1)^3(-itanA)^5 + 28(1)^2(-itanA)^6 + 8(1)(-itanA)^7 + (-itanA)^8

and
(1+ itan@)^8 = 2sec^8@cos8@ - (1-itan@)^8

Expanding (1-itan@)^8 in Bernoulli, real parts only occur when we have even powers of i

RE{1-itan@)^8 = - {1 - 28tan^2@ + 70tan^4@ - 28tan^6@ + tan^8@}

= - {1 -28(3 - 2sqrt(2)) + 70(17 - 12sqrt(2)) - 28(99 - 70sqrt(2)) + 577 - 408sqrt(2)}
= - {[-83+1190-2772+577] + [-56-840+1960-408]sqrt2)}
= 1088 - (768*sqrt2)
= 1088 - 1086.116015902536997479696940193

using
tan(pi/8) = sqrt(2) - 1
tan^2(pi/8) = 3 - 2sqrt(2)
tan^4(pi/8) = 17 - 12sqrt(2)
tan^6(pi/8) = 99 - 70sqrt(2)
tan^8(pi/8) = 577 - 408sqrt(2)

and taking real parts,

RE{(1+ itan@)^8} = 2sec^8@cos8@ + RE{ - (1-itan@)^8}
= -2{4-2sqrt(2)}^4 + 1088 - (768*sqrt2) = 1.8839840974631
= 64(12*sqrt2 - 17)

using sec(pi/8) = sqrt{4-2sqrt(2)}
sec^8(pi/8) = {4-2sqrt(2)}^4
cos(pi) = -1

QED

4. Originally Posted by tsunamijon
A more fun way

By Bernoulli expansion,

(1-itanA))^8 = (1)^8 + 8(1)^7(-itanA) + 28(1)^6(-itanA)^2 + 56(1)^5(-itanA)^3 + 70(1)^4(-itanA)^4 + 56(1)^3(-itanA)^5 + 28(1)^2(-itanA)^6 + 8(1)(-itanA)^7 + (-itanA)^8

and
(1+ itan@)^8 = 2sec^8@cos8@ - (1-itan@)^8

Expanding (1-itan@)^8 in Bernoulli, real parts only occur when we have even powers of i

RE{1-itan@)^8 = - {1 - 28tan^2@ + 70tan^4@ - 28tan^6@ + tan^8@}

= - {1 -28(3 - 2sqrt(2)) + 70(17 - 12sqrt(2)) - 28(99 - 70sqrt(2)) + 577 - 408sqrt(2)}
= - {[-83+1190-2772+577] + [-56-840+1960-408]sqrt2)}
= 1088 - (768*sqrt2)
= 1088 - 1086.116015902536997479696940193

using
tan(pi/8) = sqrt(2) - 1
tan^2(pi/8) = 3 - 2sqrt(2)
tan^4(pi/8) = 17 - 12sqrt(2)
tan^6(pi/8) = 99 - 70sqrt(2)
tan^8(pi/8) = 577 - 408sqrt(2)

and taking real parts,

RE{(1+ itan@)^8} = 2sec^8@cos8@ + RE{ - (1-itan@)^8}
= -2{4-2sqrt(2)}^4 + 1088 - (768*sqrt2) = 1.8839840974631
= 64(12*sqrt2 - 17)

using sec(pi/8) = sqrt{4-2sqrt(2)}
sec^8(pi/8) = {4-2sqrt(2)}^4
cos(pi) = -1

QED
Yuck

We have from what I said earlier:

$\text{re} (1+i \tan(\pi/8))^8=(\sec(\pi/8))^8 \cos(\pi)=$ $-\frac{1}{(\cos(\pi/8))^8}$

Now we use the double angle formula for $\cos$ to reduce this to an expression in $\cos(\pi/4)=1/\sqrt{2}$ which we simplify to find the given result.

CB