How do you use the double angle formula to rewrite $5 - 10 {sin}^{2} x$ $5-10sin^2x$ ?

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Answer 1 · 2015-04-26T17:32:59

Think about (list) double angle formulas until you come to one that looks useful.

$sin (2 θ) = 2 sin θ cos θ$ $sin(2 theta) = 2sin theta cos theta$ Doesn't look helpful

$cos (2 θ) = {cos}^{2} θ - {sin}^{2} θ$ $cos(2 theta) = cos^2 theta - sin^2 theta$ -- Hmmmm, probably not but there are other forms for this one

$cos (2 θ) = 1 - 2 {sin}^{2} θ$ $cos(2 theta) = 1 - 2sin^2 theta$ . Compare this to what we started with.

If I factored out a 5, I'd have:

$5 - 10 {sin}^{2} x = 5 (1 - 2 {sin}^{2} x)$ $5-10sin^2x = 5(1-2sin^2x)$ and that's just

$5 cos (2 x)$ $5cos(2x)$

So there it is.

$5 - 10 {sin}^{2} x = 5 cos (2 x)$ $5-10sin^2x = 5cos(2x)$

How do you use the double angle formula to rewrite 5-10sin^2x5−10sin2x5-10sin^2x?