How do you use the sum and double angle identities to find sin3x?

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Answer 1 · 2015-04-26T17:58:57

Since the angle sum formula of sinus:

$sin (α + β) = sin α cos β + cos α sin β$ $sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta$ ,

and the double angle formula of cosine:

$cos 2 α = {cos}^{2} α - {sin}^{2} α = 2 {cos}^{2} α - 1 = 1 - 2 {sin}^{2} α$ $cos2alpha=cos^2alpha-sin^2alpha=2cos^2alpha-1=1-2sin^2alpha$

than:

$sin 3 x = sin (2 x + x) = sin 2 x cos x + cos 2 x sin x =$ $sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=$

$= 2 sin x cos x \cdot cos x + (1 - 2 {sin}^{2} x) sin x =$ $=2sinxcosx*cosx+(1-2sin^2x)sinx=$

$= 2 sin x {cos}^{2} x + sin x - 2 {sin}^{3} x =$ $=2sinxcos^2x+sinx-2sin^3x=$

$= 2 sin x (1 - {sin}^{2} x) + sin x - 2 {sin}^{3} x =$ $=2sinx(1-sin^2x)+sinx-2sin^3x=$

$= 2 sin x - 2 {sin}^{3} x + sin x - 2 {sin}^{3} x =$ $=2sinx-2sin^3x+sinx-2sin^3x=$

$= 3 sin x - 4 {sin}^{3} x$ $=3sinx-4sin^3x$ .