prove: cos 4A = cos^4 A - 6 cos^2 A sin ^2 A + sin^4 A?

Question

"A", represents the angle

tried to solve it but I got "8 cos 4A" instead of "cos 4A". Pls... help.

Answer 1

cos 4A = cos^2 (2A) - sin^2 (2A)
= (cos^2 A - sin^2 A)^2 - (2 sin A cos A)^2
= cos^4 A + sin^4 A - 2 cos^2 A sin^2 A - 4 sin^2 A cos^2 A
= cos^4 A - 6 cos^2 A sin^2 A + sin^4 A

using the identities
cos 2x = cos^2 x - sin^2 x
sin 2x = 2 sin x cos x

Answer 2

Prove the identity
cos 4A = cos^4 A - 6 cos²A sin²A + sin^4 A

Let's work with the left hand side.

Left Hand Side = cos 4A = cos²2A - sin²2A

= (cos²A - sin²A)² - (2sinA cosA)²

= cos^4 A - 2cos²A sin²A + sin^4 A - 4cos²A sin²A

= cos^4 A - 6 cos²A sin²A + sin^4 A = Right Hand Side
__________

Answer 3

It looks like the right side of the equation is a quadratic. Try factoring it then using a Cosine Trig Identity to solve.

Answer 4

Cos 2A = cos (A+A) = cosAcosA – sinAsinA = cos^2(A) – sin^2(A) ---------(a)
Sin 2A = sin (A+A) = sinAcosA + cosAsinA = 2sinAcosA ---------(b)
Following the syntax of line (a);
Cos 4A = cos (2A+2A) = Cos^2(2A) – Sin^2(2A) ---------(c)
substituting for Cos2A and Sin2A in line (c), we have;
Cos 4A = (Cos^2(A) – Sin^2(A))^2 – (2SinACosA)^2 ----------(d)
expanding (Cos^2(A) – Sin^2(A))^2 in line (d) gives;
(Cos^2(A) – Sin^2(A))^2 = Cos ^4(A) – 2Sin^2(A)Cos^29A) + Sin^4(A) ------------(e)
expanding (2SinACosA)^2 in line (d) gives;
(2SinACosA)^2 = 4Sin^2(A)Cos^2(A) -----------(f)
substituting lines (e) & (f) in line (d), we have;
Cos 4A = Cos ^4(A) – 2Sin2^(A)Cos^2(A) + Sin^4(A) – 4Sin^2(A)Cos^2(A) ------(g)
Expanding we have;
Cos 4A = Cos^4(A) – 6Sin^2(A)Cos^2(A) + Sin^4(A)

Q.E.D

Answer 5

cos4A = cos2(2A) =

(cos2A)^2 - sin(2A))^2=

[(cosA)^2-(sinA)^2]^2 - [ 2sinAcosA]^2=

(cosA)^4 -2(sinA)^2(cosA)^2+(sinA)^4 - 4(sinA)^2(cosA)^2

=(cosA)^4 -6(sinA)^2(cosA)^2 + (sinA)^4