To show that the equation \( 4 \sin(A) \cos^3(A) - 4 \cos(A) \sin^3(A) = \sin(4A) \) holds true, let's go through the following trigonometric identities and steps.
1. Double Angle Identities:
[tex]\[
\sin(2A) = 2 \sin(A) \cos(A)
\][/tex]
[tex]\[
\cos(2A) = \cos^2(A) - \sin^2(A)
\][/tex]
2. Expressing \(\sin(4A)\):
We know:
[tex]\[
\sin(4A) = 2 \sin(2A) \cos(2A)
\][/tex]
Since \(\sin(2A) = 2 \sin(A) \cos(A)\) and \(\cos(2A) = \cos^2(A) - \sin^2(A)\), substitute these into the equation:
[tex]\[
\sin(4A) = 2 \left(2 \sin(A) \cos(A)\right) \left(\cos^2(A) - \sin^2(A)\right)
\][/tex]
Simplify this:
[tex]\[
\sin(4A) = 4 \sin(A) \cos(A) \left(\cos^2(A) - \sin^2(A)\right)
\][/tex]
3. Simplifying Further:
Let's expand the terms:
[tex]\[
\sin(4A) = 4 \sin(A) \left(\cos^3(A) - \cos(A) \sin^2(A)\right)
\][/tex]
Distribute \(\sin(A)\):
[tex]\[
\sin(4A) = 4 \sin(A) \cos^3(A) - 4 \sin(A) \cos(A) \sin^2(A)
\][/tex]
[tex]\[
\sin(4A) = 4 \sin(A) \cos^3(A) - 4 \cos(A) \sin^3(A)
\][/tex]
Hence, we have shown that:
[tex]\[
4 \sin(A) \cos^3(A) - 4 \cos(A) \sin^3(A) = \sin(4A)
\][/tex]
Thus, the equation [tex]\( 4 \sin(A) \cos^3(A) - 4 \cos(A) \sin^3(A) = \sin(4A) \)[/tex] holds true.
