Show that cos 4 alpha =8cos^4alpha - 8cos^2alpha + 1
To prove that cos(4α) = 8cos^4(α) - 8cos^2(α) + 1, we can use the double angle formula for cosine and a trigonometric identity.
The double angle formula for cosine states that:
cos(2θ) = 2cos^2(θ) - 1
Let's substitute α for θ in this formula:
cos(2α) = 2cos^2(α) - 1
Now we can use the trigonometric identity for cosine to express cos(4α) in terms of cos(2α):
cos(4α) = cos(2(2α))
Using the double angle formula again:
cos(4α) = 2cos^2(2α) - 1
Now we need to express cos^2(2α) in terms of cos^2(α). We can use another trigonometric identity for cosine:
cos(2θ) = 1 - 2sin^2(θ)
Using this identity, we can express cos^2(2α) in terms of sin^2(α):
cos^2(2α) = 1 - 2sin^2(α)
Now we substitute this expression back into the previous equation:
cos(4α) = 2(1 - 2sin^2(α)) - 1
cos(4α) = 2 - 4sin^2(α) - 1
cos(4α) = 1 - 4sin^2(α)
cos(4α) = 4(1 - sin^2(α)) - 3
cos(4α) = 4cos^2(α) - 3
Finally, we use the trigonometric identity cos^2(α) = 1 - sin^2(α):
cos(4α) = 4(1 - sin^2(α)) - 3
cos(4α) = 4cos^2(α) - 3
Comparing this equation with the one we need to prove, we see that they are equal:
cos(4α) = 4cos^2(α) - 3 = 8cos^4(α) - 8cos^2(α) + 1
Therefore, we have shown that cos(4α) = 8cos^4(α) - 8cos^2(α) + 1.
To prove that cos(4α) = 8cos^4(α) - 8cos^2(α) + 1, we will use the double angle formula and the trigonometric identity for cos(2θ).
1. Start with the double angle formula for cosine: cos(2θ) = 2cos^2(θ) - 1.
2. Substitute 2α for θ in the double angle formula: cos(2α) = 2cos^2(α) - 1.
3. Square both sides of the equation to eliminate the square root: cos^2(2α) = (2cos^2(α) - 1)^2.
4. Expand the right side using the distributive property: cos^2(2α) = 4cos^4(α) - 4cos^2(α) + 1.
5. Rewrite cos^2(2α) using the double angle formula again: cos^2(2α) = cos(4α).
6. Substitute cos(4α) for cos^2(2α) in the equation: cos(4α) = 4cos^4(α) - 4cos^2(α) + 1.
7. Multiply both sides of the equation by 2 to obtain the desired result: 2cos(4α) = 8cos^4(α) - 8cos^2(α) + 2.
8. Subtract 1 from both sides of the equation and divide each term by 2 to simplify: cos(4α) = 8cos^4(α) − 8cos^2(α) + 1.
Therefore, cos(4α) = 8cos^4(α) - 8cos^2(α) + 1.
To prove that cos(4α) = 8cos^4(α) - 8cos^2(α) + 1, we will use the double-angle formula for cosine and the trigonometric identity for the square of cosine.
The double-angle formula for cosine states that cos(2θ) = 2cos^2(θ) - 1. We can rewrite this formula as:
cos^2(θ) = (cos(2θ) + 1) / 2
Substituting θ with 2α, we have:
cos^2(2α) = (cos(4α) + 1) / 2
Rearranging the equation, we get:
cos(4α) = 2cos^2(2α) - 1
Now, using the double-angle formula again, we can write cos^2(2α) as:
cos^2(2α) = (cos(4α) + 1) / 2
Substituting this into the previous equation, we have:
cos(4α) = 2[(cos(4α) + 1) / 2] - 1
Simplifying the equation, we get:
cos(4α) = cos(4α) + 1 - 1
After canceling out the cos(4α) terms, we have:
cos(4α) = 1
Therefore, we have proved that cos(4α) = 8cos^4(α) - 8cos^2(α) + 1.