prove:
2sinx+sin2x = 2sin^3(x)/1-cosx
LS = 2sinx + 2sinxcosx
= 2sinx(1 + cosx)
= 2sinx(1 + cosx)(1-cosx)/(1-cosx)
= 2sinx(1 - cos^2x)/(1-cosx)
= 2sinx(sin^2x)/(1-cosx)
= 2sin^3x/(1-cosx)
= RS
Thank you Reiny!
To prove the equation 2sin(x) + sin(2x) = 2sin^3(x)/(1-cos(x)), we need to simplify both sides of the equation using trigonometric identities. Let's break it down step by step:
Starting with the left side:
1. We know that sin(2x) = 2sin(x)cos(x) from the double angle formula.
2. Substituting sin(2x) = 2sin(x)cos(x) into the equation, we get:
2sin(x) + 2sin(x)cos(x)
Now, let's simplify the right side:
1. We can rewrite sin^3(x) as (sin^2(x))sin(x).
2. Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can substitute sin^2(x) = 1 - cos^2(x).
3. Replacing sin^2(x) with 1 - cos^2(x), we have:
2(1 - cos^2(x))sin(x)/(1-cos(x))
Now, let's focus on simplifying each side separately:
Left side: 2sin(x) + 2sin(x)cos(x)
Right side: 2(1 - cos^2(x))sin(x)/(1-cos(x))
= 2sin(x)(1 - cos^2(x))/(1 - cos(x))
= 2sin(x)(1 - cos(x))(1 + cos(x))/(1 - cos(x))
= 2sin(x)(1 + cos(x))
As we can see, both sides are equal. Therefore, we have proved that 2sin(x) + sin(2x) = 2sin^3(x)/(1-cos(x)).