how can you prove the following identity: Sin(2x) - Cot (x) = - Cot(x)Cos(2x) please help me!!!!?

Answer 1

I'll start with the right side of the identity and show it equals the left side.

-cot(x)cos(2x) = -[cos(x)/sin(x)][1 - 2sin^2(x)]
= -cos(x)/sin(x) + 2cos(x)sin(x) = -cot(x) + sin(2x)
= sin(2x) - cot(x)

Answer 2

Sin(2x) - Cot (x) = - Cot(x)Cos(2x)
2sin(x)cos(x) - cos(x)/sin(x) =[ -cos(x)/sin(x)](cos²(x) - sin²(x))
2sin²(x)cos(x) - cos(x) = -cos(x)(cos²(x) - sin²(x))
-2sin²(x) + 1 = cos²(x) - sin²(x)
1 = cos²(x) + sin²(x)

QED

Answer 3

i'd write each side as powers of e. for instance: sin(x)= (e^(ix)-e^(-ix))/2

then, it should be easy to show their equal.