sin3A = 3 sinx cos^2(A) - sin^3(A)
= 3sinA(1-sin^2(A)) - sin^3(A)
= 3sinA - 4sin^3(A)
clearly,
x/2 = A
x = 2A
Sin3A=3sinA-4sin(x/2)^3
find x interms of A?
PLZ show
working got no ideal at all
3 answers
How did you come ot x/2=A?
Back in the 'good ol' days' we had to memorize or have at hand many trig identities. One of them was the expansion
sin(3A) =3sinA - 4 sin^3 A , which Steve actually developed for you.
Compare that with your given equation.
All the terms are the same except the last term, where we have
4sin^3 (x/2) vs -4sin^3 A
What has replaced your x/2 ??
picking up Steve's lines ....
clearly,
x/2 = A
x = 2A
sin(3A) =3sinA - 4 sin^3 A , which Steve actually developed for you.
Compare that with your given equation.
All the terms are the same except the last term, where we have
4sin^3 (x/2) vs -4sin^3 A
What has replaced your x/2 ??
picking up Steve's lines ....
clearly,
x/2 = A
x = 2A
1 answer