Question

hi pls help me how to prove the following.

Answer 1

\[\cos ^{4}\theta=\frac{ 1 }{ 8 }\cos 4\theta +\frac{ 1 }{ 2 }\cos 2\theta +\frac{ 3 }{ 8 }\]

Answer 2

I rmemeber reading about this somewhere and they first started witht e fact that cos^4thetais (cos^2)^2 and the fact that cos^2=1-2sin^2theta and thats all I can rmemebr sorry lol

Answer 3

I will try..stay here

Answer 4

cos^4 x
= (cos^2 x) (cos^2 x)
= (cos^2 x) (1 - sin^2 x)
= (cos^2 x - sin^2 x cos^2 x)
= 2/2 * (cos^2 x - sin^2 x cos^2 x)
= 1/2 * (2cos^2 x - 2sin^2 x cos^2 x)
= 1/2 * (cos^2 x + cos^2 x - 2sin^2 x cos^2 x)
= 1/2 * (1 - sin^2 x + cos^2 x - 2sin^2 x cos^2 x)
= 4/8 * (1 - sin^2 x + cos^2 x - 2sin^2 x cos^2 x)
= 1/8 * (4 - 4sin^2 x + 4cos^2 x - 8sin^2 x cos^2 x)
= 1/8 * (3 + 1 - 4sin^2 x + 4cos^2 x - 8sin^2 x cos^2 x)
= 1/8 * (3 + 1^2 - 4sin^2 x + 4cos^2 x - 8sin^2 x cos^2 x)
= 1/8 * (3 + (sin^2 x + cos^2 x)^2 - 4sin^2 x + 4cos^2 x - 8sin^2 x cos^2 x)

= 1/8 * (3 + sin^4 x + cos^4 x + 2sin^2 x cos^2 x - 4sin^2 x + 4cos^2 x - 8sin^2 x cos^2 x)

= 1/8 * (3 + sin^4 x + cos^4 x - 4sin^2 x + 4cos^2 x - 6sin^2 x cos^2 x)

= 1/8 * (3 + sin^4 x + cos^4 x - 4sin^2 x + 4cos^2 x - 2sin^2 x cos^2 x - 4sin^2 x cos^2 x)

= 1/8 * (3 + sin^4 x + cos^4 x - 2sin^2 x cos^2 x - 4sin^2 x + 4cos^2 x - 4sin^2 x cos^2 x)

= 1/8 * (3 + (cos^2 x - sin^2 x)^2 - 4sin^2 x + 4cos^2 x - (2sin x cos x)^2)

= 1/8 * (3 + (cos 2x)^2 - 4sin^2 x + 4cos^2 x - (sin 2x)^2)
= 1/8 * (3 + cos^2 2x - 4sin^2 x + 4cos^2 x - sin^2 2x)
= 1/8 * (3 + 4cos^2 x - 4sin^2 x + cos^2 2x - sin^2 2x)
= 1/8 * (3 + 4(cos^2 x - sin^2 x) + cos^2 2x - sin^2 2x)
= 1/8 * (3 + 4(cos 2x) + (cos^2 2x - sin^2 2x))
= 1/8 * (3 + 4cos 2x + (cos 4x))
= (1/8) (3 + 4cos 2x + cos 4x)

Answer 5

please excuse if its wrong..not completely mine

Answer 6

I assume you have to do many of these and as I can;t fully understand this textbook with its proof, I'll just provide it, I mean you kinda learn by looking at this answer anyways but herre's the proof, if its more or less the same as