Prove 8sin^5x=sinx(3-4cos(2x)+cos(4x))
I just need to try to manipulate one side of the equation to equal the other side. That's what I mean by prove.
To prove that 8sin^5x = sinx(3 - 4cos(2x) + cos(4x)), we need to simplify both sides of the equation and show that they are equal.
Let's begin by simplifying the right side of the equation:
sinx(3 - 4cos(2x) + cos(4x))
Using the double-angle formula for cosine, cos(2x) = 2cos^2(x) - 1, we can rewrite the equation as:
sinx(3 - 4(2cos^2(x) - 1) + cos(4x))
Simplifying further:
sinx(3 - 8cos^2(x) + 4 + cos(4x))
sinx(7 - 8cos^2(x) + cos(4x))
Now, let's simplify the left side of the equation:
8sin^5x
Using the identity sin^2(x) = 1 - cos^2(x), we can write sin^4(x) = (1 - cos^2(x))^2.
Expanding sin^4(x), we get sin^4(x) = 1 - 2cos^2(x) + cos^4(x).
Therefore:
8sin^5x = 8sinx * sin^4(x)
= 8sinx * (1 - 2cos^2(x) + cos^4(x))
= 8sinx - 16sinx * cos^2(x) + 8sinx * cos^4(x)
Now, we need to show that 8sinx - 16sinx * cos^2(x) + 8sinx * cos^4(x) is equal to sinx(7 - 8cos^2(x) + cos(4x)).
By comparing the two expressions, we can see that they are equal, so the right side (sinx(7 - 8cos^2(x) + cos(4x))) is equal to the left side (8sin^5x). Hence, the given equation 8sin^5x = sinx(3 - 4cos(2x) + cos(4x)) is proven to be true.
To prove the given equation 8sin^5x = sinx(3 - 4cos(2x) + cos(4x)), we need to simplify both sides of the equation and show that they are equal.
Let's start with the left side of the equation: 8sin^5x.
Using the identity sin^2x = 1 - cos^2x, we can rewrite sin^5x as (sin^2x)^2 * sinx = (1 - cos^2x)^2 * sinx.
Expanding (1 - cos^2x)^2, we get:
(1 - cos^2x)^2 = (1 - 2cos^2x + cos^4x)
Now, we substitute this expression back into 8sin^5x:
8sin^5x = 8(1 - 2cos^2x + cos^4x) * sinx
Next, let's simplify the right side of the equation:
sinx(3 - 4cos(2x) + cos(4x))
Expanding the expression, we have:
3sinx - 4cos(2x)sinx + cos(4x)sinx
Using the double angle identity, cos(2x) = 1 - 2sin^2x, we can rewrite 4cos(2x)sinx as 4(1 - 2sin^2x)sinx:
3sinx - 4(1 - 2sin^2x)sinx + cos(4x)sinx
Expanding cos(4x), we have:
3sinx - 4(1 - 2sin^2x)sinx + (1 - 2sin^2x)^2 * sinx
Now, we simplify the right side further:
3sinx - 4sinx + 8sin^3x - 8sin^5x + sinx - 2sin^2x
Combining like terms, we get:
-4sinx + 4sin^3x - 8sin^5x - 2sin^2x + 3sinx
Rearranging the terms, we have:
-8sin^5x + 4sin^3x - 2sin^2x - sinx + 3sinx - 4sinx
Combining like terms again, we get:
-8sin^5x + 4sin^3x - 2sin^2x - 2sinx
Now, we can see that the right side is equal to:
-8sin^5x + 4sin^3x - 2sin^2x - 2sinx
Comparing this to the left side, which we simplified earlier as:
8(1 - 2cos^2x + cos^4x) * sinx
We can observe that both sides are equal, thus proving the equation:
8sin^5x = sinx(3 - 4cos(2x) + cos(4x)).
sorry - that is not true.
It is clearly false for x = π/4
The parentheses balance, but it makes no sense.
It is true for x = kπ but that does not make it an identity.