simplify 1-cos2x = (sin2x)(tanx)?

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answers:
MyRank say: 1-cos2x = (sin2x) (tanx)

1-cos2x = (2sinxcosx) (sinx/cosx)

1-cos2x = (2sin²x)

1-cos2x = (2(1-cos²x))

1-cos2x = 2-2cos²x

1-(2cos²x-1) = 2-2cos²x

1-2cos²x+1 = 2-2cos²x

0 = 0.
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Como say: 1 - ( 1 - 2 sin²x ) = 2 sin²x / cos x
2 sin²x = 2 sin²x / cos x
sin²x cos x - sin²x = 0
sin²x ( cos x - 1 ) = 0
sin x = 0 , cos x = 1
x = 0° , 180° , 360°
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la console say: Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = x

cos(x + x) = cos(x).cos(x) - sin(x).sin(x)

cos(2x) = cos²(x) - sin²(x) ← memorize this result as (1)



Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = b = x

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2.sin(x).cos(x) ← memorize this result as (2)



= 1 - cos(2x) → recall (1)

= 1 - [cos²(x) - sin²(x)]

= 1 - cos²(x) + sin²(x) → recall: cos²(x) + sin²(x) = 1

= [cos²(x) + sin²(x)] - cos(²x) + sin²(x)

= cos²(x) + sin²(x) - cos(²x) + sin²(x)

= 2.sin²(x)

= 2.sin²(x) * 1

= 2.sin²(x) * [cos(x)/cos(x)]

= 2.sin(x).sin(x) * [cos(x)/cos(x)]

= 2.sin(x).cos(x).[sin(x)/cos(x)] → you know that: sin(x)/cos(x) = tan(x)

= 2.sin(x).cos(x).tan(x) → recall (2)

= [2.sin(x).cos(x)].tan(x) → recall (2)

= sin(2x).tan(x)
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Johnathan say: Let's see...at first, I would assume you got to prove if this is an identity.

sin(2x)tan(x) (right)

= 2 sin(x)cos(x)(sin(x)/cos(x))

= 2 sin^2(x)

= sin^2(x) + sin^2(x)

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

= 1 - (cos^2(x) - sin^2(x))

= 1 - cos(2x) (left). Done.

I was right. This is an identity -- if cos(x) is nonzero (x =/= n(pi) + pi/2 if n is an integer) since on the right side you have tan(x).
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Vaman say: It is correct. You try it.
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