Exact Values Using Double-Angle Formulae

To find $\sin 2\theta$ and $\cos 2\theta$ from $\sin\theta$ alone: find $\cos\theta$ first (right triangle or Pythagoras), then apply the double-angle formulae. Keep everything as exact fractions.

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find the exact values of $\sin 2\theta$ and $\cos 2\theta$. [5]

The working

Step 1, find $\cos\theta$. With $\sin\theta = \frac{3}{5}$, a $3$–$4$–$5$ right triangle gives the adjacent side $4$, and since $\theta$ is acute $\cos\theta$ is positive: $\cos\theta = \frac{4}{5} \quad \text{(M1, A1)}$

Step 2, apply $\sin 2\theta = 2\sin\theta\cos\theta$: $\sin 2\theta = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25} \quad \text{(M1, A1)}$

Step 3, apply $\cos 2\theta = 1 - 2\sin^2\theta$: $\cos 2\theta = 1 - 2\left(\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25} \quad \text{(A1)}$

(Equivalently $\cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.)

Where the marks are won and lost

• Get $\cos\theta$ with the right sign for the quadrant. Here $\theta$ is acute, so $\cos\theta = +\frac45$; in another quadrant it could be negative.
• Keep exact fractions throughout, the question says “exact”, so a decimal like $0.96$ won’t score the accuracy mark.
• Any correct form of the $\cos 2\theta$ formula works; pick the one that uses what you already have.

Common mistakes

• Taking $\cos\theta$ negative when $\theta$ is acute.
• Writing $\sin 2\theta = 2\sin\theta$ (forgetting the $\cos\theta$).
• Using $\cos 2\theta = 1 - 2\cos^2\theta$ (wrong; it’s $2\cos^2\theta - 1$ or $1 - 2\sin^2\theta$).

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