Trig identities

Question

(2sin^2(x))/1-cosx=2sinx+sin2x

Answer 1

2sin²x/(1 - cosx)= 2(1 - cosx)(1 + cosx)/(1 - cosx) = 2(1 + cosx)

Your identity wrong(if x = 90° we get 0 = 2

Answer 2

2sin^2(x)/(1-cosx) = 2sinx +sin2x

check `1st to see if it's really an identity

by plugging in a couple values for x

such as x=0

cos0=1 so that makes the left side undefinied, but the right side =0

it's not an identity.

try another value x=60 degrees

left side = 3, right side = 2sqr3

It's definitely not an identity.

just one counterexample is enought to prove it's not an identity

an identity means its true for all values of x.

this saves a lot of time hopelessly trying to prove it's an identity

as well as avoiding a mistake where you sometines think you've proved it

(you may have typed 2 when you meant 3 as the exponent of sinx, for the 2sin^2(x) on the left side of the problem)

Answer 3

2 sin x + sin 2x =

2 sin x + 2 sin x cos x =

2 sin x ( 1 + cos x) =

2 sin x ( 1 + cos x) ( 1 - cos x) / (1 - cos x) =

2 sin³ x / (1 - cos x)

pretty close to your 2 sin² x / (1 - cos x), but not the same. Perhaps, check your post for typos(?)