Graphs of sine, cosine and tangent, the identities $\sin^2\theta+\cos^2\theta=1$ and $\tan\theta=\sin\theta/\cos\theta$, solving trig equations, and the sine and cosine rules.

What this dot point is asking

WJEC wants you to recognise and sketch the graphs of $\sin$, $\cos$ and $\tan$, to use the identities $\sin^2\theta + \cos^2\theta = 1$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, to solve trigonometric equations in a given interval, and to solve non-right-angled triangles with the sine rule, the cosine rule and the area formula. Trigonometry recurs in calculus and in the A2 trig identities, so the foundations here matter.

The answer

The trig graphs

You should be able to sketch each graph and read symmetry and periodicity from it:

• $y = \sin\theta$: period $360^{\circ}$, range $[-1, 1]$, odd, passes through the origin.
• $y = \cos\theta$: period $360^{\circ}$, range $[-1, 1]$, even, starts at $(0, 1)$.
• $y = \tan\theta$: period $180^{\circ}$, range all reals, asymptotes at $\theta = 90^{\circ}, 270^{\circ}, \ldots$

Identities

The Pythagorean identity is the workhorse: it converts $\sin^2$ into $1 - \cos^2$ (or vice versa) so an equation can be written entirely in one ratio.

Solving trigonometric equations

The sine and cosine rules

For a triangle with sides $a, b, c$ opposite angles $A, B, C$:

Use the sine rule when you have a side and its opposite angle plus one more piece; use the cosine rule when you have two sides and the included angle, or all three sides.

Examples in context

Example 1. Bearings and the cosine rule. A ship sails $12\,\text{km}$ then turns through an interior angle of $110^{\circ}$ and sails $9\,\text{km}$. The direct distance back is $c$ where $c^2 = 12^2 + 9^2 - 2(12)(9)\cos 110^{\circ} = 144 + 81 + 73.9 = 298.9$, so $c \approx 17.3\,\text{km}$. The cosine rule handles the non-right triangle that a real journey produces.

Example 2. Area to find an angle. A triangle with sides $6\,\text{cm}$ and $10\,\text{cm}$ has area $15\,\text{cm}^2$. Using $\text{Area} = \tfrac{1}{2}ab\sin C$, $15 = \tfrac{1}{2}(6)(10)\sin C$, so $\sin C = 0.5$ and $C = 30^{\circ}$ (or the obtuse $150^{\circ}$). The area formula works backwards to an angle.

Try this

Q1. Solve $\tan\theta = 1$ for $0^{\circ} \le \theta \le 360^{\circ}$. [2 marks]

• Cue. Tangent is positive in quadrants 1 and 3, so $\theta = 45^{\circ}, 225^{\circ}$.

Q2. In triangle $ABC$, $A = 40^{\circ}$, $B = 75^{\circ}$ and $a = 9\,\text{cm}$. Find $b$. [3 marks]

• Cue. Sine rule: $b = \dfrac{9\sin 75^{\circ}}{\sin 40^{\circ}} \approx 13.5\,\text{cm}$.

Q3. Show that $\dfrac{1 - \cos^2\theta}{\sin\theta} = \sin\theta$. [2 marks]

• Cue. The numerator is $\sin^2\theta$ by the identity, so the fraction is $\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta$.