Question

Question #74046

Answer 1

Proof by De moivres theorem....

Explanation:

We can prove the following statement by use of de moivres theorem

`color(red)(cos^4 theta -= 1/8 ( cos4theta + 4cos2theta + 3), AA theta`

`AA theta` - Means "for all theta"

We can use `color(blue)(omega = costheta +isintheta`

Now using de moivres theorm we yield:

`omega + 1/omega = 2costheta`

and...

`omega^n + 1/omega^n = 2cosn theta`

`=> ( 2costheta)^4 = (omega + 1/omega)^4`

`=> 16cos^4 theta = (omega + 1/omega)^4`

Now using the binomial expansion

`=> 16cos^4 theta = omega^4 + (4omega^3 * 1/omega) + (6omega^2 * 1/omega^2) + (4omega*1/omega^3) + 1/omega^4`

`=> 16cos^4 theta = omega^4 + 1/omega^4 + 4(omega^2 + 1/omega^2 ) + 6`

Using `omega^n + 1/omega^n = 2cosntheta`

`=> 16cos^4 theta = 2cos4theta + 8cos2theta + 6`

`=> cos^4 theta = 1/16 ( 2cos4theta + 8cos2theta + 6 )`

`color(red)( =>cos^4theta = 1/8 ( cos4theta + 4cos2theta + 3 )`

Answer 2

Proof by conventional trigonometric identities...

Explanation:

We can prove the follwing statment by trig identities...

`color(red)( cos^4 theta -= 1/8 ( cos4theta + 4cos2theta + 3 ) ,AA theta`

`AA theta` - Meaning "For all theta "

We can `color(blue) ( cos^2 theta - sin^2 theta -= cos2theta`

`=> 2cos^2 theta - 1 -= cos2theta`

We can use this identity...

`=> cos^2 theta -= 1/2(cos2theta + 1 )`

`cos^4 theta = (cos^2 theta)^2`

`=> cos^4 theta = (1/2 ( cos2theta +1 ) ) ^2`

`=> cos^4 theta = 1/4( cos^2 2theta +2cos2theta +1 )`

Again we can use the fact that `color(red)(cos^2 theta = 1/2(cos2theta +1)`

Adapting this...

`=> color(blue)(cos^2 2theta = 1/2 (cos4theta +1 )`

`"Hence this: " cos^4 theta = 1/4( color(blue)(cos^2 2theta) +2cos2theta +1 )`

`"Becomes this: " cos^4 theta = 1/4 ( color(blue)(1/2(cos4theta +1)) +2cos2theta +1 )`

`=> cos^4 theta -= 1/4 ( 1/2cos4theta + 2cos2theta +3/2 )`

`=> cos^4 theta -= 1/4 * 1/2(cos4theta + 4cos2theta +3 )`

`color(red)( => cos^4 theta -= 1/8 ( cos4theta + 4cos2theta + 3 )`