If 2Cosθ = x+1/x So, Prove That ? Cos2θ = 1/2(x²+1/x²)

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If 2Cosθ = x+1/x So, Prove That ? Cos2θ = 1/2(x²+1/x²)

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Answer 1 · 2018-05-20T15:50:36

$L H S = cos 2 θ$ $LHS=cos2theta$

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

$= 2 {(\frac{1}{2} (x + \frac{1}{x}))}^{2} - 1$ $=2(1/2(x+1/x))^2-1$

$= \frac{1}{2} (x^{2} + \frac{1}{x^{2}} + 2 \cdot x \cdot \frac{1}{x}) - 1$ $=1/2(x^2+1/x^2+2*x*1/x)-1$

$= \frac{1}{2} (x^{2} + \frac{1}{x^{2}}) + \frac{1}{2} \cdot 2 - 1$ $=1/2(x^2+1/x^2)+1/2*2-1$

$= \frac{1}{2} (x^{2} + \frac{1}{x^{2}}) = R H S$ $=1/2(x^2+1/x^2)=RHS$

Answer 2 · 2018-05-20T16:13:10

Sqaring the given equation we get
$cos (θ) = \frac{1}{4} \cdot (x^{2} + \frac{1}{x^{2}} + 2)$ $cos(theta)=1/4*(x^2+1/x^2+2)$ then by multiplying by $2$ $2$ we get
$2 {cos}^{2} (θ) = \frac{1}{2} \cdot (x^{2} + \frac{1}{x^{2}}) + 1$ $2cos^2(theta)=1/2*(x^2+1/x^2)+1$ from here we get
$2 {cos}^{2} (θ) - 1 = \frac{1}{2} (x^{2} + \frac{1}{x^{2}})$ $2cos^2(theta)-1=1/2(x^2+1/x^2)$

Explanation:

we used ${(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2$ $(x+1/x)^2=x^2+1/x^2+2$
$cos (2 x) = 2 {cos}^{2} (x) - 1$ $cos(2x)=2cos^2(x)-1$

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If 2Cosθ = x+1/x So, Prove That ? Cos2θ = 1/2(x²+1/x²)

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