I need to verify this equation and I get stuck after trying everything I can think of. Help!
cos(4x)=4[sin(x)cos(x)-sin^3(x)cos(x)]
To verify the equation cos(4x) = 4[sin(x)cos(x) - sin^3(x)cos(x)], we can use trigonometric identities to simplify both sides of the equation and see if they are equal.
Let's start by simplifying the right-hand side of the equation:
sin(x)cos(x) - sin^3(x)cos(x)
We know that sin^2(x) = 1 - cos^2(x), so substituting this into the equation:
sin(x)cos(x) - sin^3(x)cos(x)
= sin(x)cos(x) - (1 - cos^2(x))sin(x)cos(x)
= sin(x)cos(x) - sin(x)cos(x) + cos^3(x)sin(x)
= cos^3(x)sin(x)
Now, let's simplify the left-hand side of the equation:
cos(4x)
We can express cos(4x) using the double-angle identities:
cos(4x)
= cos(2x + 2x)
= cos^2(2x) - sin^2(2x)
Using the double-angle identities again, we can express cos^2(2x) and sin^2(2x) in terms of cos(2x) and sin(2x):
cos^2(2x) - sin^2(2x)
= (cos(2x) + sin(2x))(cos(2x) - sin(2x))
By using the trigonometric identity cos^2(x) - sin^2(x) = cos(2x), we can simplify further:
(cos(2x) + sin(2x))(cos(2x) - sin(2x))
= (cos(2x) + sin(2x))(cos(2x) - sin(2x))(cos(2x) + sin(2x) + cos(2x) - sin(2x))
= (cos(2x) + sin(2x))^2 - (cos(2x) - sin(2x))^2
= cos^2(2x) + 2cos(2x)sin(2x) + sin^2(2x) - (cos^2(2x) - 2cos(2x)sin(2x) + sin^2(2x))
= 4cos(2x)sin(2x)
Now, comparing the simplified left-hand side and right-hand side of the equation, we have:
4cos(2x)sin(2x) = cos^3(x)sin(x)
Since we want to verify that the equation is true, we need to show that these two expressions are equivalent for all values of x.
To proceed further, we can use the double-angle identities for sine and cosine:
cos(2x) = 2cos^2(x) - 1
sin(2x) = 2sin(x)cos(x)
Substituting these identities into the equation, we have:
4(2cos^2(x) - 1)(2sin(x)cos(x)) = cos^3(x)sin(x)
Expanding and simplifying both sides, we get:
4(4cos^3(x)sin(x) - 2cos(x)sin(x)) = cos^3(x)sin(x)
16cos^3(x)sin(x) - 8cos(x)sin(x) = cos^3(x)sin(x)
16cos^3(x)sin(x) - 9cos(x)sin(x) = 0
sin(x)cos(x)(16cos^2(x) - 9) = 0
Therefore, either sin(x) = 0, cos(x) = 0, or 16cos^2(x) - 9 = 0.
If sin(x) = 0, it means x = 0 or x = π.
If cos(x) = 0, it means x = π/2 or x = 3π/2.
If 16cos^2(x) - 9 = 0, solving for cos(x), we get cos(x) = ±√(9/16) = ±3/4. This leads to four possible solutions: x = π/3, x = -π/3, x = 5π/3, and x = -5π/3.
These values of x satisfy the equation cos(4x) = 4[sin(x)cos(x) - sin^3(x)cos(x)]. You can verify this by substituting them back into the original equation for cos(4x) and comparing both sides.
Therefore, sin(x) = 0, cos(x) = 0, cos(x) = ±3/4, or x = 0, π/2, 3π/2, π/3, -π/3, 5π/3, or -5π/3 are the solutions for which the equation is true.