How do you verify the identity $cos x - 1 = \frac{cos 2 x - 1}{2 (cos x + 1)}$ $cosx-1=(cos2x-1)/(2(cosx+1))$ ?

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How do you verify the identity $cos x - 1 = \frac{cos 2 x - 1}{2 (cos x + 1)}$ $cosx-1=(cos2x-1)/(2(cosx+1))$ ?

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Answer 1 · 2017-03-30T00:14:05

Start with the right side:

$\frac{cos 2 x - 1}{2 (cos x + 1)}$ $(cos2x-1) / (2(cosx+1))$

Use the identity $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x = color(red)(2cos^2x-1)$

$\frac{2 {cos}^{2} x - 1 - 1}{2 (cos x + 1)}$ $(color(red)(2cos^2x-1)-1) / (2(cosx+1))$

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

Now divide both the numerator and denominator by 2:

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

Factor the numerator:

$\frac{(cos x + 1) (cos x - 1)}{cos x + 1}$ $((cosx+1)(cosx - 1)) / (cosx + 1)$

$((cancel(cosx+1))(cosx - 1)) / cancel(cosx+1)$

$cosx - 1$

Which is the left side.

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How do you verify the identity $cos x - 1 = \frac{cos 2 x - 1}{2 (cos x + 1)}$ $cosx-1=(cos2x-1)/(2(cosx+1))$ ?

1 Answer

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