How do you simplify $\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x}$ $(sin^4x-cos^4x)/(sin^2x-cos^2x)$ ?

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How do you simplify $\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x}$ $(sin^4x-cos^4x)/(sin^2x-cos^2x)$ ?

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Answer 1 · 2016-09-14T13:03:14

$1$ $1$

Explanation:

$\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x} = ({sin}^{2} x + {cos}^{2} x) \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x - {cos}^{2} x} = 1$ $(sin^4x-cos^4x)/(sin^2x-cos^2x)=(sin^2x+cos^2x)(sin^2x-cos^2x)/(sin^2x-cos^2x) = 1$

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How do you simplify $\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x}$ $(sin^4x-cos^4x)/(sin^2x-cos^2x)$ ?

1 Answer

Explanation:

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How do you simplify (sin^4x-cos^4x)/(sin^2x-cos^2x)sin4x−cos4xsin2x−cos2x(sin^4x-cos^4x)/(sin^2x-cos^2x)?

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Explanation:

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How do you simplify $\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x}$ $(sin^4x-cos^4x)/(sin^2x-cos^2x)$ ?