Question

Verify:
{sinX^2/2(1-cosX)} + {sinX^2/2(1+cosX)} = 1

Answer 1

whatever i do i keep on getting 2sinX^2 = 1-cosX^2 which doesn't fit any identities.

Answer 2

\[\frac{ \sin^2x }{ 2(1-cosX) } + \frac{ \sin^2x }{ 2(1+cosX) } = 1\] is this the equation?

Answer 3

yes sir

Answer 4

Okay then well what do you want to do first, find a common denominator? or use your identities & change \[\sin^2x\] into \[\frac{ 1-\cos(2x) }{ 2 }\] ?

Answer 5

i started with the common denominator 2(1-cos^2 X)

Answer 6

so after doing that you end up with \[\frac{ 2\sin^2x }{ 2(1-\cos^2x) }\] & the 2 reduce out.

Answer 7

\[2\sin ^{2}x (1+cosx) + 2\sin ^{2}x(1-cosx)\]

Answer 8

yes that's what i got

Answer 9

wait no i got 2sin^2X / 1-cos^2x

Answer 10

Yeah, but you forgot the 2 in the denominator. so then instead of \[\frac{ \sin^2x }{ 1-\cos^2x } = 1 \rightarrow \sin^2x = 1-\cos^2x\]

Answer 11

Do you understand it now?

Answer 12

2sin^2x(1+cosx)+2sin^2x(1−cosx) this was my numerator ; 2(1-cos^2x) the denominator, so the 2's cancelled out leaving sin^2x(1+cosx) +sin^2x(1-cosx) in the numerator

Answer 13

after extending the numerator, i get 2sin^2x still

Answer 14

there must be a step i did wrong somewhere that im not seeing then

Answer 15

I'll rewrite it out step by step & then maybe that will help you find where you messed up or where we are miscommunicating.

Answer 16

thanks a lot i'm breaking my head here.

Answer 17

No problem!

Answer 18

it says 403 forbidden

Answer 19

Now try it:

Answer 20

i see where i messed up ......... i multiply all the numerators by 2... my eyes tricked me

Answer 21

thank you kindly for your time, it was a silly mistake after all.

Answer 22

No problem! & it happens after dealing with any problem for so long. aha