Math · Precalculus · Worked example

Use a double-angle identity

Given sin θ = 3/5 and π/2 < θ < π, find sin 2θ and cos 2θ.

sin θ = \frac{3}{5}

Find cos θ

From sin²θ + cos²θ = 1, cos²θ = 1 − 9/25 = 16/25. Cosine is negative in quadrant II, so cos θ = −4/5.

cos θ = - \sqrt{1 - {(\frac{3}{5})}^{2}} = - \frac{4}{5}

Use sin 2θ = 2 sin θ cos θ.

sin 2 θ = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}

\begin{array}{c} sin 2 θ = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) \\ = - \frac{24}{25} \end{array}

Any of the three forms works; 1 − 2 sin²θ uses the given value directly.

cos 2 θ = 1 - 2 {(\frac{3}{5})}^{2} = \frac{7}{25}

sin²2θ + cos²2θ must be 1, and 576/625 + 49/625 = 625/625.

{(- \frac{24}{25})}^{2} + {(\frac{7}{25})}^{2} = 1

sin 2θ = −24/25 and cos 2θ = 7/25.

Given cos θ = 5/13 and 3π/2 < θ < 2π, find sin 2θ.

Show the answer and explanation

Sine is negative in quadrant IV, so sin θ = −√(1 − 25/169) = −12/13. Then sin 2θ = 2(−12/13)(5/13) = −120/169.

\begin{array}{l} - \sqrt{1 - \frac{25}{169}} = - \frac{12}{13} \\ 2 \cdot (- \frac{12}{13}) \cdot \frac{5}{13} = - \frac{120}{169} \end{array}

In Math, change 3/5 to 5/13 and −4/5 to −12/13: the rows then give sin 2θ = −120/169 and cos 2θ = 119/169.

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Open the example inputs, change a value and keep a useful result on your board.

Your existing work stays on this device. Examples open as editable copies.