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[tex]\[ 4 \sin A \cdot \cos^3 A - 4 \cos A \cdot \sin^3 A = \sin 4A \][/tex]

Sagot :

GinnyAnswer
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.
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