Math · Precalculus · Worked example

Prove a trigonometric identity

Prove that cos⁴x − sin⁴x = cos 2x.

{cos}^{4} x - {sin}^{4} x = cos 2 x

Start from the more complicated side

The left side has fourth powers, so work on it and aim for cos 2x. It is a difference of squares: (cos²x)² − (sin²x)².

{cos}^{4} x - {sin}^{4} x = ({cos}^{2} x - {sin}^{2} x) ({cos}^{2} x + {sin}^{2} x)

\begin{aligned} {cos}^{4} x - {sin}^{4} x \\ = ({cos}^{2} x - {sin}^{2} x) \\ \cdot ({cos}^{2} x + {sin}^{2} x) \end{aligned}

The second factor is cos²x + sin²x, which is 1 for every x.

({cos}^{2} x - {sin}^{2} x) \cdot 1 = {cos}^{2} x - {sin}^{2} x

\begin{array}{c} ({cos}^{2} x - {sin}^{2} x) \cdot 1 \\ = {cos}^{2} x - {sin}^{2} x \end{array}

cos²x − sin²x is one of the three forms of cos 2x, so the left side equals the right side for every x.

{cos}^{2} x - {sin}^{2} x = cos 2 x

cos⁴x − sin⁴x = (cos²x − sin²x)(cos²x + sin²x) = cos²x − sin²x = cos 2x, so the identity holds for every x.

Prove that (1 − cos x)(1 + cos x) = sin²x.

Show the answer and explanation

(1 − cos x)(1 + cos x) is a difference of squares, 1 − cos²x. Rearranging sin²x + cos²x = 1 gives 1 − cos²x = sin²x.

\begin{array}{l} (1 - cos x) (1 + cos x) \\ = 1 - {cos}^{2} x \\ = {sin}^{2} x \end{array}

In Graph, the two sides draw a single curve. Change the right side to cos²x and a second curve separates from it: that statement is not an identity.

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