Asked by abdo
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.
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.
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