Final answer:
To prove tan(A)*(2cos³(A) - cos(A)) = sin(A) - 2sin³(A), we use trigonometric identities to simplify the left side of the equation by replacing tan(A) with sin(A)/cos(A) and using the double-angle formula and Pythagorean identity to factor and simplify the expression to match the right side.
Step-by-step explanation:
To prove the expression tan(A)*(2cos³(A) - cos(A)) = sin(A) - 2sin³(A), let's start by using trigonometric identities to simplify the left side of the equation.
Recall that tan(A) = sin(A)/cos(A). So, tan(A) can be replaced with sin(A)/cos(A), which gives us:
(sin(A)/cos(A))*(2cos³(A) - cos(A))
Distributing the sin(A)/cos(A) term inside the brackets, we get:
2sin(A)cos²(A)/cos(A) - sin(A)cos(A)/cos(A)
Simplifying the fractions, we get:
2sin(A)cos(A) - sin(A)
We know from the Pythagorean identity that cos²(A) = 1 - sin²(A), and cos(A) = √(1 - sin²(A)). We can use the double-angle formula sin(2A) = 2sin(A)cos(A) to rewrite 2sin(A)cos(A).
Thus, 2sin(A)cos(A) becomes sin(2A), which can be further simplified to 2sin(A) - sin³(A) according to the given trigonometric identities. So the original expression on the left side now looks like sin(A) - sin(A)sin²(A).
Factor out sin(A), which gives:
sin(A)(1 - sin²(A))
Use the identity sin²(A) = 1 - cos²(A) to obtain:
sin(A)(cos²(A))
This is equivalent to:
sin(A) - 2sin³(A), which proves the original expression.