Thread: GRADE 12 PRE-CAL IDENTITIES

question is:

sin(3x) = 3sinx-4sin³x

there is a clue..
it's that sin(3x)=sin (2x+x)

please help!! and as fast as possible!!
message me or whatever is possible!

sin(3x) = 3sinx-4sin³x

there is a clue..
it's that sin(3x)=sin (2x+x)

We can start by rewriting the left side of the equation as per suggestion:
sin(2x+x) = 3sinx-4sin³x

Use the following identity for sine sum formula:
sin(A+B)=sin A cos B + cos A sin B

Again, rewrite the left side of the equation with the formula:
sin2x*cosx + cos2x*sinx = 3sinx - 4sin³x

Use the following double angle formulas:
sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2

Rewrite the left side of the equation using these double angle formulas:
2sinxcosx*cosx + ((cosx)^2 - (sinx)^2)*sinx = 3sinx - 4sin³x

Divide both sides by sinx to yield:
2(cosx)^2 + (cosx)^2 - (sinx)^2 = 3 - 4*(sinx)^2

Add like terms:
3*(cosx)^2 = 3 - 3*(sinx)^2

Add sine term to both sides:
3*(cosx)^2 + 3*(sinx)^2 = 3

Divide both sides by 3 and you get the true identity:

(cosx)^2 + (sinx)^2 = 1

I hope this is useful.