sin3x=(sinx)(3-asin^2x)
Wow!
is that arcsin^2(x) ???
are we solving for x?
or is it proving the identity?
proving the identity
sin3x=(sinx)(3-sin^2x)
Your identity is wrong, it should say:
sin3x=(sinx)(3-4sin^2x)
try this:
sin(3x) = sin(2x + x)
= sin2xcosx + cos2xsinx
keep replacing in terms of sinx, it comes out pretty smoothly.
proving the identity:
sin3x=(sinx)(3-4sin^2x)
proving the identity:
sin3x=(sinx)(3-4sin^2x)
Reiny gave you a hint:
in3x=(sinx)(3-4sin^2x)
try this:
sin(3x) = sin(2x + x)
= sin2xcosx + cos2xsinx
keep replacing in terms of sinx, it comes out pretty smoothly.
To prove the identity sin3x = (sinx)(3-4sin^2x), we can start by using the double-angle identity for sine:
sin(2x) = 2sin(x)cos(x)
Let's use this identity to rewrite sin(3x):
sin(3x) = sin(2x + x)
Using the angle sum identity for sine, we have:
sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)
Now, let's express sin(2x) and cos(2x) in terms of sin(x) and cos(x) using the double-angle identities:
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x)
Substituting these expressions back into our previous equation, we get:
sin(2x + x) = 2sin(x)cos(x)cos(x) + (1 - 2sin^2(x))sin(x)
= 2sin(x)cos^2(x) + sin(x) - 2sin^3(x)
Next, we can factor out sin(x) from the first two terms:
sin(2x + x) = sin(x)(2cos^2(x) + 1) - 2sin^3(x)
Now, we can simplify the expression in the parentheses:
2cos^2(x) + 1 = 2(1 - sin^2(x)) + 1
= 2 - 2sin^2(x) + 1
= 3 - 2sin^2(x)
Substituting this back into our equation, we have:
sin(2x + x) = sin(x)(3 - 2sin^2(x)) - 2sin^3(x)
As you can see, this matches the right side of the given identity (sinx)(3-4sin^2x). Hence, we have proven the identity sin3x = (sinx)(3-4sin^2x).