Sin3A=3sinA-4sin(x/2)^3
find x interms of A?
PLZ show
working got no ideal at all
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
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
To find x in terms of A, we need to simplify the equation and isolate x.
Let's start by simplifying the equation step by step:
1. Sin 3A = 3sin A - 4sin^3(x/2)
2. Using the trigonometric identity for sin 3A, we have:
3sin A - 4sin^3(x/2) = 3sin A - 4[3sin(x/2) - 4sin^3(x/2)]
3. Distributing the -4 to the terms inside the brackets:
3sin A - 4[3sin(x/2) - 4sin^3(x/2)] = 3sin A - 12sin(x/2) + 16sin^3(x/2)
4. Combining like terms and rearranging:
3sin A - 12sin(x/2) + 16sin^3(x/2) = 3sin A + 16sin^3(x/2) - 12sin(x/2)
Now, we can equate the terms on both sides of the equation:
3sin A - 12sin(x/2) + 16sin^3(x/2) = 3sin A + 16sin^3(x/2) - 12sin(x/2)
Since the terms on both sides are the same, we can cancel them out:
3sin A - 12sin(x/2) + 16sin^3(x/2) - (3sin A + 16sin^3(x/2) - 12sin(x/2)) = 0
Simplifying further:
3sin A - 12sin(x/2) + 16sin^3(x/2) - 3sin A - 16sin^3(x/2) + 12sin(x/2) = 0
Now, we can combine like terms:
-12sin(x/2) + 12sin(x/2) = 0
Since the equation simplifies to 0 = 0, this indicates that the equation is identity and holds true for any value of x and A.
Therefore, the original equation Sin 3A = 3sin A - 4sin^3(x/2) does not specifically relate x and A.