how can you confirm the identity
cos^4x = (1/8)(3+ 4cos2x+ cos4x)
and
sin4x = (4sinxcosx)(2cos^2x-1)
You use the identity:
exp(ix) = cos(x) + i sin(x)
You can then write:
cos(x) = [exp(ix)+ exp(-ix)]/2
Take the fourth power:
cos^4(x) =
1/16 [exp(4 i x) + 4 exp(2 i x) + 6 +
4 exp(-2 i x) + exp(-4 i x)] =
1/8 {3 + [exp(4 i x) + exp(-4 i x)]/2 + 4 [exp(2 i x) + exp(-2 i x)]/2} =
1/8 [3 + cos(4 x) + 4 cos(2 x)]
sin(4x) can be expanded in the usual way by using the doubling formulas:
sin(4x) = 2 sin(2x) cos(2x) =
4 sin(x) cos(x)cos(2x)=
4 sin(x) cos(x) [2 cos^2(x) - 1]
The general method also works, but it slightly more laborious in this simple case:
You start with
exp(ix) = cos(x) + i sin(x)
and take the fourth power of both sides and take the imaginary part:
Im[exp(4 i x)] = 4 cos^3(x)sin(x) -
4 cos(x) sin^3(x)
Im[exp(4 i x)] = sin(4 x), so we have:
sin(4 x) =
4 cos(x)sin(x)[cos^2(x) -sin^2(x)]
Using that sin^2(x) = 1 - cos^2(x) gives you the desired result.
To confirm the identity of the given equations, we need to simplify both sides of the equation separately and show that they are equal to each other.
Let's start with the first equation: cos^4x = (1/8)(3 + 4cos2x + cos4x).
To simplify the left-hand side (LHS), we'll expand the power using the identity: cos²x = (1 + cos2x)/2.
So cos^4x = (cos^2x)^2 = [(1 + cos2x)/2]^2.
Expanding the square, we get cos^4x = [(1 + cos2x)^2/(2^2)].
Simplifying further, cos^4x = (1 + 2cos2x + cos^2(2x))/4.
Now, let's simplify the right-hand side (RHS): (1/8)(3 + 4cos2x + cos4x).
We can start by multiplying everything inside the parentheses by (1/8):
RHS = (1/8)(3/8 + 4cos2x/8 + cos4x/8).
Simplifying further, RHS = (3/64 + cos2x/16 + cos4x/64).
To compare the two sides, we need to find a common denominator for both expressions.
The common denominator is 64, so we will need to multiply the numerator and the denominator of (1 + cos2x)^2 by 16 to have the same denominator.
cos^4x = [(1 + cos2x)^2/(2^2)] = [(16 + 16cos2x + 16cos^2(2x))/64].
The LHS is now in terms of a common denominator of 64.
Comparing the simplified LHS and RHS expressions, we can observe that they are the same.
Therefore, we can confirm that cos^4x = (1/8)(3 + 4cos2x + cos4x) is an identity.
Now let's move on to the second equation: sin4x = (4sinxcosx)(2cos^2x - 1).
To simplify the RHS, we'll start by using the double-angle identity for sine: sin2x = 2sinxcosx.
The RHS can be rewritten as: (4sinxcosx)(2cos^2x - 1) = 2(2sinxcosx)(2cos^2x - 1).
Expanding further, we have: 2(2sinxcosx)(2cos^2x - 1) = 4sinxcosx(4cos^2x - 2).
Next, we can simplify the expression: 4sinxcosx(4cos^2x - 2) = 8sinxcos^3x - 4sinxcosx.
Now, let's simplify the LHS of the equation: sin4x.
Using the double-angle identity for sine once again, sin4x = 2sin2xcos2x.
Substituting 2sin2x for sin4x, we have: 2sin2xcos2x = 2(2sinxcosx)(2cos^2x).
Simplifying further, we get: 2(2sinxcosx)(2cos^2x) = 8sinxcos^3x.
Comparing the simplified LHS and RHS expressions, we can observe they are equal.
Therefore, we can confirm that sin4x = (4sinxcosx)(2cos^2x - 1) is an identity.
By performing step-by-step simplifications on both sides of each equation and proving their equality, we have confirmed the identities.