Question

((1-cosx)^2 +sin^2x)/1-cosx = 2 ...... wtf? can some explain this please

Answer 1

you're dealing with this right?

\[\Large \frac{(1-\cos(x))^2 + \sin^2(x)}{1-\cos(x)} = 2\]

Answer 2

i know sin^2x = 1-cos^2x and (1-cosx)^2 should be 1+ cos^2x right? i dont see how the res of the algebra works out to = 2

Answer 3

yes

Answer 4

so let's use sin^2x = 1-cos^2x

Answer 5

\[\Large \frac{(1-\cos(x))^2 + \sin^2(x)}{1-\cos(x)} = 2\]
\[\Large \frac{(1-\cos(x))^2 + 1-\cos^2(x)}{1-\cos(x)} = 2\]
\[\Large \frac{(1-\cos(x))(1-\cos(x)) + 1-\cos^2(x)}{1-\cos(x)} = 2\]
\[\Large \frac{1(1-\cos(x))-\cos(x)(1-\cos(x)) + 1-\cos^2(x)}{1-\cos(x)} = 2\]
\[\Large \frac{1-\cos(x)-\cos(x)+\cos^2(x) + 1-\cos^2(x)}{1-\cos(x)} = 2\]
\[\Large \frac{2-2\cos(x)}{1-\cos(x)} = 2\]
Hopefully you see how to finish up?

Answer 6

yeah i see it thanks!!

Answer 7

you're welcome

Answer 8

actually i dont see that either. i thought i did, but when i tried to do it i get 2-2cos x-2cos x-3cos. @jim_thompson5910

Answer 9

which part are you stuck at again?

Answer 10

finishing it

Answer 11

so you understand all my steps shown above? or no?

Answer 12

yeah i understand that

Answer 13

so you then would factor 2 from the numerator

Answer 14

and then cancel the 1-cos(x) terms

Answer 15

wow, that shouldnt have been a aha moment, i dont see why i didnt see that. thanks again

Answer 16

that's ok