CRAY
Region: GB
Thursday 18 June 2026 14:05:36 GMT
216141
28755
128
6630
Music
Download
Comments
D🤓 :
this is further away from maths
2026-06-18 21:12:16
5035
chanel :
Bro just wanted to use his Apple Pencil
2026-06-19 07:20:26
2974
🇺🇦🇱🇹 :
is this what no more exams does to people
2026-06-18 19:09:38
1581
silverwish :
I was genuinely believing you
2026-06-19 01:55:29
1427
Aisha Jasmine :
I think you’ve been very bored since a levels finished
2026-06-18 15:07:43
1109
isa :
ok socrates 😍
2026-06-18 15:05:58
517
sofia❣️ :
My English teacher would love you😍😍
2026-06-18 16:27:35
496
sonyawww (SAW MCR) :
this is 100% how i am moving cus i take fm and english lit
2026-06-18 21:16:44
462
CRAY :
wait guys did i eat x
2026-06-18 14:10:24
367
jana 😹 :
bull and shit 💕💕
2026-06-19 23:35:18
297
twice and ive- aot is peak :
I don’t take further maths but you’re right
2026-06-18 18:14:08
195
frankcastlespistol :
Allat for three marks btw
2026-06-18 19:54:45
93
cody 🕷️ :
Im dying 😭😭
2026-06-18 14:14:50
57
:
🤣 mind you this is the easy bit
2026-06-18 19:19:35
33
𝔱𝔬𝔯𝔦𝔪𝔢 (björk’s version) :
I never heard of Renee Sánse?? Xx
2026-06-19 17:52:38
31
MONARCH :
"3 marks"😭😭😭
2026-06-19 06:18:35
25
🔵 Dylan 🔵 :
This is not difficult... 😭
2026-06-18 18:44:04
21
aj ✨🤍 :
so first apply de moivres theorum for z^n: since z = cos theta + i sin theta the theory tells us that for any integer (n) that z^n = (cos theta + i sin theta)^n = cos(n theta) + i sin(n theta) then using the laws of indicies 1/z^n = z^-n and applying de moivres theorem for negetives z^-n = (cos theta + i sin theta) ^-n = cos(-n theta) + i sin (-n theta) then using the even-odd trigonometric identities cos(-x) = cos(x) and sin(-x) = - sin(x) that simplifys to 1/z^n = cos(n theta) - i sin(n theta) then to substitute the expressions into the equation so z^n - 1/z^n = [cos(n theta) + i sin(n theta)] - [cos(n theta) - i sin(n theta)] then distribute the negetive and collect like terms z^n - 1/z^n = cos(n theta) - cos(n theta) + i sin(n theta) + i sin(n theta) so z^n - 1/z^n = 2 i sin(n theta) which proves z^n + 1/z^n = 2 cos(n theta)
2026-06-19 11:58:39
20
trevinotv :
I’m pissing myself
2026-06-20 11:22:25
18
mudasxr 🇦🇫 :
women in stem 🥰
2026-06-18 14:54:34
17
AH♠️ :
This is arguably the easiest fm topic
2026-06-18 19:43:39
16
To see more videos from user @crayante, please go to the Tikwm
homepage.
