Respuesta :

Answer:

see explanation

Step-by-step explanation:

Using the double angle identity for cosine

cos2x = 2cos²x - 1

Given

cos( [tex]\frac{0}{2}[/tex] ) = [tex]\frac{1}{2}[/tex]( p + [tex]\frac{1}{p}[/tex] ) , then

cosΘ = 2[ [tex]\frac{1}{2}[/tex](p + [tex]\frac{1}{p}[/tex] ) ]² - 1

         = 2 [ [tex]\frac{1}{4}[/tex](p² + 2 + [tex]\frac{1}{p^{2} }[/tex] ) ] - 1  ← distribute by 2

         = [tex]\frac{1}{2}[/tex](p² + 2 + [tex]\frac{1}{p^{2} }[/tex] ) - 1 ← distribute by [tex]\frac{1}{2}[/tex]

         = [tex]\frac{1}{2}[/tex] p² + 1 + [tex]\frac{1}{2p^2}[/tex] - 1

        = [tex]\frac{1}{2}[/tex] p² + [tex]\frac{1}{2p^2}[/tex] ← factor out [tex]\frac{1}{2}[/tex] from each term

        = [tex]\frac{1}{2}[/tex] ( p² + [tex]\frac{1}{p^{2} }[/tex] ) ← as required

Alfpfeu

Answer:

Step-by-step explanation:

Hello, please consider the following.

[tex]cos(\theta)=2cos^2(\dfrac{\theta}{2})-1\\\\=2\left(\dfrac{p+\dfrac{1}{p}}{2}\right)^2-1\\\\=\dfrac{p^2+\dfrac{1}{p^2}+2}{2}-1\\\\=\dfrac{1}{2}(p^2+\dfrac{1}{p^2})+1-1\\\\=\dfrac{1}{2}(p^2+\dfrac{1}{p^2})[/tex]

Thank you

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