Question

Using sin(A+B) = SinAcosB+ CosAsinB and cos(A+B)= CosAcosB-sinAsinB. Show that tan(2A)= 2tanA/(1-TanAtanB)
2. Sin3A= 3sinA-4(sinA^3)

Answer 1

First derive the tangent addition formula.

From
sin(A+B)=sinA cosB + cosA sinB
and
cos(A+B)=cosA cosB − sinA sinB,
divide numerator and denominator by cosA cosB:

tan(A+B) = sin(A+B)/cos(A+B)
= (sinA cosB + cosA sinB)/(cosA cosB − sinA sinB)
= ( (sinA/cosA) + (sinB/cosB) ) / ( 1 − (sinA/cosA)(sinB/cosB) )
= (tanA + tanB)/(1 − tanA tanB).

Now set B = A to get tan(2A):
tan(2A) = (tanA + tanA)/(1 − tanA·tanA) = 2 tanA/(1 − tan^2 A).

Next prove sin3A.

Write sin3A = sin(2A + A) = sin2A cosA + cos2A sinA.
Use sin2A = 2 sinA cosA and cos2A = 1 − 2 sin^2 A (or cos^2A − sin^2A):

sin3A = (2 sinA cosA)cosA + (1 − 2 sin^2 A) sinA
= 2 sinA cos^2 A + sinA − 2 sin^3 A
= sinA(2(1 − sin^2 A) + 1 − 2 sin^2 A)
= sinA(3 − 4 sin^2 A)
= 3 sinA − 4 sin^3 A.

Hence
tan(2A) = 2 tanA/(1 − tan^2 A)
and
sin3A = 3 sinA − 4 sin^3 A.