Question

Indicate the correct alternatives for the following-:
(listing the question and option)

Answer 1

\[for 0<\phi<\pi/2\]

Answer 2

if \[x=\sum_{n=0}^{\infty}\cos^{2n} \phi,y=\sum_{n=0}^{\infty}\sin^{2n}\phi \]

Answer 3

\[z=\sum_{n=0}^{\infty}\cos ^{2n}\phi(\sin^{2n}\phi)\]

Answer 4

then-:(more then one correct)
xyz=xz+y
xyz=xy+z
xyz=x+y+z
xyz=yz+x

Answer 5

@ganeshie8 @wio @ParthKohli pls help

Answer 6

@Luigi0210

Answer 7

@ganeshie8 help

Answer 8

by 2n , do you mean power or derivative ?

Answer 9

My first idea is to convert it into exponential form and see what happens.

Answer 10

@Kainui still not sure 2n is power or n derevative ?

Answer 11

I think if it was derivatives then it wouldn't converge.

Answer 12

it would if it was double derevative then for sin note that
sin x=- sin '' x
cos x = cos '' x
ect

Answer 13

Yeah but those sums look like:

1-1+1-1+1-1+... and won't converge

Answer 14

Answer my messages

Answer 15

Actually if you look at this like a geometric series, you can solve this fairly easily it would seem. For instance, \[\Large x=\frac{1}{1-\cos^2(x)}\]

The rest should follow from trig identities fairly nicely.

Answer 16

\[\large xyz = \dfrac{1}{\sin^2\phi \cos^2\phi \left(1-\sin^2\phi \cos^2\phi\right)}\]

?

Answer 17

nice :)

Answer 18

@taylor12344 answer my messages toooo

Answer 19

does any1 now about Vn method shortcut

Answer 20

above is a shortcut method ^

Answer 21

yes i got the question

Answer 22

you just need to know the infinite sum of converging geometric series formula : \(\dfrac{1}{1-r}\)

Answer 23

i was able to solve this question on my own

Answer 24

but i forgot to close the question

Answer 25

what method did u use

Answer 26

i used sum of infinity in a/1-r and then i started substiuting the value of

Answer 27

{sorry} i used the value of phi

Answer 28

and ticked the suitble option

Answer 29

great, we all are on the same page :)

Answer 30

ok

Answer 31

It's really not a hard formula to derive. In fact, it's probably one of the most satisfying things in math to do.

\[\Large S=1+a+a^2+a^3+... \\ \Large S-1=a+a^2+a^3+... \\ \Large \frac{S-1}{a}= 1+a+a^2+a^3+... \\ \Large \frac{S-1}{a}=S\] Now you can just solve for your sum S with a little algebra. =P

Answer 32

yes i solved the question