How do you prove ${sin}^{4} θ - {cos}^{4} θ = {sin}^{2} θ - {cos}^{2} θ$ $sin^4theta-cos^4theta=sin^2theta-cos^2theta$ ?

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Answer 1 · 2016-10-05T19:06:17

${sin}^{4} θ - {cos}^{4} θ = {sin}^{2} θ - {cos}^{2} θ$ $sin^4 theta-cos^4 theta = sin^2 theta-cos^2 theta$

Left Side: $= {sin}^{4} θ - {cos}^{4} θ$ $=sin^4 theta-cos^4 theta$

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

$= ({sin}^{2} θ - {cos}^{2} θ) \cdot 1$ $=(sin^2theta-cos^2theta)*1$

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

$=$ $=$ Right Side

How do you prove sin4θ−cos4θ=sin2θ−cos2θsin^4theta-cos^4theta=sin^2theta-cos^2theta?