How do you express $cos (4 θ)$ $cos(4theta)$ in terms of $cos (2 θ)$ $cos(2theta)$ ?

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Answer 1 · 2015-06-10T23:05:38

$cos (4 θ) = 2 {(cos (2 θ))}^{2} - 1$ $cos(4theta) = 2(cos(2theta))^2-1$

Start by replacing $4 θ$ $4theta$ with $2 θ + 2 θ$ $2theta+2theta$

$cos (4 θ) = cos (2 θ + 2 θ)$ $cos(4theta) = cos(2theta+2theta)$

Knowing that $cos (a + b) = cos (a) cos (b) - sin (a) sin (b)$ $cos(a+b) = cos(a)cos(b)-sin(a)sin(b)$ then

$cos (2 θ + 2 θ) = {(cos (2 θ))}^{2} - {(sin (2 θ))}^{2}$ $cos(2theta+2theta) = (cos(2theta))^2-(sin(2theta))^2$

Knowing that ${(cos (x))}^{2} + {(sin (x))}^{2} = 1$ $(cos(x))^2+(sin(x))^2 = 1$ then

${(sin (x))}^{2} = 1 - {(cos (x))}^{2}$ $(sin(x))^2 = 1-(cos(x))^2$

$\to cos (4 θ) = {(cos (2 θ))}^{2} - (1 - {(cos (2 θ))}^{2})$ $rarr cos(4theta) = (cos(2theta))^2-(1-(cos(2theta))^2)$

$= 2 {(cos (2 θ))}^{2} - 1$ $= 2(cos(2theta))^2-1$

How do you express cos(4θ)cos(4theta) in terms of cos(2θ)cos(2theta)?