Exact Trig Values from a Given Ratio

This is a Paper 1 favourite because it is non-calculator and tests two ideas together: the identity $\sin^2\theta + \cos^2\theta = 1$, and the sign rule for quadrants. The word “obtuse” is not decoration, it decides the sign of your answers.

Given that $\sin\theta = \dfrac{5}{13}$ and that $\theta$ is obtuse, find the exact values of $\cos\theta$ and $\tan\theta$. [4]

The working

Step 1, use the Pythagorean identity to get $\cos^2\theta$: $\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \quad \text{(M1)}$

Step 2, square-root, then fix the sign. $\theta$ is obtuse, so it lies in the second quadrant, where cosine is negative: $\cos\theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13} \quad \text{(A1)}$

Step 3, tangent is sine over cosine: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\tfrac{5}{13}}{-\tfrac{12}{13}} = -\frac{5}{12} \quad \text{(M1, A1)}$

A quick sanity check: in the second quadrant only sine is positive, so cosine and tangent should both come out negative, and they do.

Where the marks are won and lost

• Exact means fractions, not decimals. $-\frac{12}{13}$ scores; $-0.923$ risks the accuracy mark on a “find the exact value” question.
• The sign is a mark. $+\frac{12}{13}$ is the answer to a different question (acute $\theta$). The obtuse condition forces the negative root.
• $\tan\theta$ inherits the sign automatically once $\cos\theta$ is negative, no need to guess it separately.

Common mistakes

• Giving $\cos\theta = \frac{12}{13}$ by ignoring “obtuse”.
• Converting to decimals when exact values are demanded.
• Slipping on the $\frac{5}{13} \div \frac{12}{13}$ step (the $13$s cancel to leave $\frac{5}{12}$).

Full method: Exact Values & the Unit Circle notes. Topic home: Trigonometry pillar.