Question

x=a(b-sinb),y=a(1-cos b), prove dy/dx=cot(b/2)

Answer 1

dy/db = asinb
dx/db = a(1-cosb)

helps ?

Answer 2

another hint :
now you can divide dy/db by dx/db
and get dy/dx !

Answer 3

ill help more:
( dy/db ) / (dx/db) = asinb/ a(1-cosb) = sinb / (1-cosb) = dy/dx

Answer 4

now it is just trigo.

Answer 5

if you dont like the idea of dividing dy/db by dx/db
you can as well use the chain rule :
dy/dx = dy/db * db/dx

and you can find db/dx by isolating b from the equation with x and finding db/dx.

Answer 6

the answer i got is different from what to prove

Answer 7

you get
dy/dx = sinb / (1-cosb)

Answer 8

now it can be shown that it is the same as cot(b/2)

Answer 9

hint : sin(b) = sin(2 * b/2)
cos(b) = cos(2*b/2)
use the formulas for double angle now.

Answer 10

yes i get that
but how do we solve that to get cot(b/2)

Answer 11

look at my last response it should help you

Answer 12

sin(b) = 2sin(b/2)cos(b/2)
cos(b) = cos^2(b/2) - sin^2(b/2)

so

sinb / (1-cosb) = 2sin(b/2)cos(b/2) / (1- cos^2(b/2) + sin^2(b/2))

Answer 13

i hope you can do it from here.

Answer 14

@minatsukisaya95 are you there ?

Answer 15

yes
thank u
i got it

Answer 16

yw