Question

Please help me!!!!!!!!​

Answer 1

Explanation:

Hello, please consider the following.

`cos(\theta)=2cos^2((\theta)/(2))-1\\\\=2\left((p+(1)/(p))/(2)\right)^2-1\\\\=(p^2+(1)/(p^2)+2)/(2)-1\\\\=(1)/(2)(p^2+(1)/(p^2))+1-1\\\\=(1)/(2)(p^2+(1)/(p^2))`

Thank you

Answer 2

see explanation

Explanation:

Using the double angle identity for cosine

cos2x = 2cos²x - 1

Given

cos(
`(0)/(2)` ) =
`(1)/(2)`( p +
`(1)/(p)` ) , then

cosΘ = 2[
`(1)/(2)`(p +
`(1)/(p)` ) ]² - 1

= 2 [
`(1)/(4)`(p² + 2 +
`(1)/(p^(2) )` ) ] - 1 ← distribute by 2

=
`(1)/(2)`(p² + 2 +
`(1)/(p^(2) )` ) - 1 ← distribute by
`(1)/(2)`

=
`(1)/(2)` p² + 1 +
`(1)/(2p^2)` - 1

=
`(1)/(2)` p² +
`(1)/(2p^2)` ← factor out
`(1)/(2)` from each term

=
`(1)/(2)` ( p² +
`(1)/(p^(2) )` ) ← as required