Show that Cos 4¤=8cos^4¤ - 8cos^2 + 1
The formula that we can use is the double-angle formula, which states that cos(2x) = 2cos^2(x) - 1. We can use this formula twice to get the cos(4x) in terms of cos(x).
First, apply the double-angle formula for x:
cos(2x) = 2cos^2(x) - 1 ...(1)
Then, apply the double-angle formula again, but now for 2x:
cos(4x) = 2cos^2(2x) - 1 ...(2)
Substituting equation (1) into equation (2)
cos(4x) = 2[2cos^2(x) - 1]^2 - 1
= 2[4cos^4(x) - 4cos^2(x) + 1] - 1
= 8cos^4(x) - 8cos^2(x) + 2 - 1
= 8cos^4(x) - 8cos^2(x) + 1
Thus, cos(4x) = 8cos^4(x) - 8cos^2(x) + 1.
To prove the equation cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1, we will use the double-angle formula for cosine.
The double-angle formula for cosine states that cos(2θ) = 2cos^2(θ) - 1.
Let's start by writing the left side of the equation:
cos(4θ)
Using the double-angle formula, we can rewrite it as follows:
cos(2(2θ))
Since 2(2θ) is equal to 4θ, we get:
cos(4θ)
Now let's apply the double-angle formula to cos(2θ):
cos(2θ) = 2cos^2(θ) - 1
To calculate cos(4θ), we can substitute 2θ with (2θ):
cos(4θ) = 2cos^2(2θ) - 1
Now let's substitute (2θ) with (2θ):
cos(4θ) = 2cos^2(4θ) - 1
To simplify this further, let's substitute cos^2(4θ) with (cos^2(2θ))^2:
cos(4θ) = 2(cos^2(2θ))^2 - 1
Now we have an expression that includes cos^2(2θ). Let's substitute it with (2cos^2(θ) - 1):
cos(4θ) = 2((2cos^2(θ) - 1))^2 - 1
Expanding the square and simplifying, we get:
cos(4θ) = [2(4cos^4(θ) - 4cos^2(θ) + 1)] - 1
cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 2 - 1
cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1
Therefore, we have shown that cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1.
To prove the identity
cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1
we can start with the double angle formula, which states that:
cos(2θ) = 2cos^2(θ) - 1
Let's rewrite the expression on the right side of the given equation:
8cos^4(θ) - 8cos^2(θ) + 1
We can separate it into two parts using the identity:
cos(2θ) = 2cos^2(θ) - 1
First, let's substitute cos(2θ) into the equation:
8cos^4(θ) - 8cos^2(θ) + 1 = 8(2cos^2(θ) - 1)^2 - 8cos^2(θ) + 1
Next, let's expand the squared term using the binomial formula:
8(2cos^2(θ) - 1)^2 - 8cos^2(θ) + 1 = 8(4cos^4(θ) - 4cos^2(θ) + 1) - 8cos^2(θ) + 1
Now, let's distribute the 8:
8(4cos^4(θ) - 4cos^2(θ) + 1) - 8cos^2(θ) + 1 = 32cos^4(θ) - 32cos^2(θ) + 8 - 8cos^2(θ) + 1
Simplifying further:
32cos^4(θ) - 32cos^2(θ) + 8 - 8cos^2(θ) + 1 = 32cos^4(θ) - 40cos^2(θ) + 9
Now, we need to show that this expression is equal to cos(4θ). Using a trigonometric identity, we can rewrite cos(4θ) as follows:
cos(4θ) = cos^2(2θ) - sin^2(2θ)
Using the double angle formula again, we have:
cos^2(2θ) - sin^2(2θ) = cos^2(2θ) - (1 - cos^2(2θ))
Simplifying:
cos^2(2θ) - (1 - cos^2(2θ)) = 2cos^2(2θ) - 1
Now, substituting 2θ with θ, we have:
2cos^2(θ) - 1
Therefore, we have shown that:
cos(4θ) = 2cos^2(θ) - 1
Comparing this with our previous expression:
32cos^4(θ) - 40cos^2(θ) + 9 = 2cos^2(θ) - 1
We have established the identity:
32cos^4(θ) - 40cos^2(θ) + 9 = 2cos^2(θ) - 1
Hence, we have proven the given identity:
cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1.