prove that cos^4 x=(1/8) 3 + 4cos2 x+ cos 4 x using binomial expansion
just recall that
cos4x = 2cos^2(2x)-1
cos2x = 2cos^2(x)-1
expand and it all works out.
To prove that cos^4(x) = (1/8)3 + 4cos^2(x) + cos(4x) using binomial expansion, we can apply the binomial theorem. The binomial theorem states that for any real number a and b, and any positive integer n, the expansion of (a + b)^n can be obtained using the formula:
(a + b)^n = C(n, 0)a^n * b^0 + C(n, 1)a^(n-1) * b^1 + C(n, 2)a^(n-2) * b^2 + ... + C(n, n-1)a^1 * b^(n-1) + C(n, n)a^0 * b^n
where C(n, k) is the binomial coefficient, given by the formula:
C(n, k) = n! / (k!(n-k)!)
Let's apply the binomial expansion to the expression cos^4(x):
cos^4(x) = (cos^2(x))^2
Let a = cos^2(x) and b = 1, so that a + b = cos^2(x) + 1.
Now we can rewrite cos^4(x) as (a + b)^2:
cos^4(x) = (a + b)^2
Using the binomial theorem formula, we expand (a + b)^2:
cos^4(x) = C(2, 0)a^2 * b^0 + C(2, 1)a^1 * b^1 + C(2, 2)a^0 * b^2
cos^4(x) = C(2, 0)a^2 + C(2, 1)a * b + C(2, 2)b^2
Simplifying the binomial coefficients:
cos^4(x) = (1)(a^2) + (2)(a)(b) + (1)(b^2)
cos^4(x) = a^2 + 2ab + b^2
Since a = cos^2(x) and b = 1, substituting back:
cos^4(x) = (cos^2(x))^2 + 2(cos^2(x))(1) + 1^2
cos^4(x) = cos^4(x) + 2cos^2(x) + 1
Now, we need to prove that cos^4(x) = (1/8)3 + 4cos^2(x) + cos(4x).
First, simplify (1/8)3:
(1/8)3 = 1/8
Substituting this in:
cos^4(x) = 1/8 + 4cos^2(x) + cos(4x)
Finally, we have proved that cos^4(x) = (1/8)3 + 4cos^2(x) + cos(4x) using binomial expansion.