Trigonometric ratios

Step 1: Is it a right-angled triangle?

If yes, use SOHCAHTOA: $\sin\theta = O/H$, $\cos\theta = A/H$, $\tan\theta = O/A$. Or use Pythagoras for the third side.

Step 2: If non-right-angled, what information do you have?

Given	Rule
2 sides + included angle	Cosine rule (find third side)
3 sides	Cosine rule rearranged (find any angle)
1 side + 2 angles	Sine rule (find any side)
2 sides + non-included angle	Sine rule (find angle) — beware ambiguous case
2 sides + included angle	Area formula $\tfrac{1}{2}ab\sin C$
3 sides (and want area)	Heron's formula (extension, beyond P1) OR cosine rule then $\tfrac{1}{2}ab\sin C$

Mnemonic. SSS and SAS → cosine rule (S for sides, A for angle). ASA and AAS → sine rule. SSA → sine rule but watch ambiguous case.

Why? The cosine rule needs three sides OR two sides plus the included angle to fix the triangle. The sine rule needs a paired angle-and-opposite-side, plus one more piece.

Sine rule (in the IAL formula booklet).

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}.$

Each side is OPPOSITE its named angle.

Use 1: Find an unknown side. Given $a, A, B$ (or any side-angle pair plus another angle):

$b = \dfrac{a \sin B}{\sin A}.$

Use 2: Find an unknown angle. Given $a, A, b$ (or similar):

$\sin B = \dfrac{b \sin A}{a}.$

Then $B = \sin^{-1}(\ldots)$ — BUT check for the ambiguous case.

Worked example. $PQ = 8$, $PR = 11$, $\angle PQR = 80°$. Find $\angle PRQ$.

• Angle $R$ is opposite side $PQ = 8$. Angle $Q = 80°$ is opposite side $PR = 11$.
• Sine rule: $\sin R / 8 = \sin 80° / 11$.
• $\sin R = 8 \sin 80° / 11 \approx 0.7162$.
• $R \approx 45.7°$.

Sketch the triangle. Always draw a labelled diagram. It makes pairing angle ↔ opposite side obvious and reveals when the ambiguous case might bite.

Edexcel tip. Mark scheme awards a method mark just for the sine rule statement; don't skip it. Write the full rule first, then your equation.

Cosine rule (in the formula booklet).

$a^{2} = b^{2} + c^{2} - 2bc \cos A,$

where $A$ is the angle OPPOSITE side $a$, AND $A$ is the angle BETWEEN sides $b$ and $c$.

Use 1: Find a side (SAS). Given two sides $b, c$ and the included angle $A$, find $a$:

$a = \sqrt{b^{2} + c^{2} - 2bc \cos A}.$

Example. $b = 9$, $c = 7$, $A = 65°$: $a^{2} = 81 + 49 - 126 \cos 65° \approx 130 - 53.25 = 76.75 \Rightarrow a \approx 8.76.$

Use 2: Find an angle (SSS). Given all three sides, rearrange:

$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}.$

The angle $A$ is opposite side $a$.

Example. $a = 5, b = 7, c = 8$: $\cos A = (49 + 64 - 25)/(2 \cdot 7 \cdot 8) = 88/112 = 0.7857 \Rightarrow A \approx 38.2°$.

Cosine rule generalises Pythagoras. When $A = 90°$, $\cos A = 0$, and the formula becomes $a^{2} = b^{2} + c^{2}$ — Pythagoras.

Edexcel tip. Calculator must be in DEGREE mode. Test: $\cos 60°$ should give $0.5$. If you get $-0.952$, you're in radians.

Area formula (in the formula booklet).

$\text{Area} = \tfrac{1}{2} ab \sin C.$

Conditions. $C$ must be the included angle — the one BETWEEN sides $a$ and $b$.

Worked example. Sides 6 and 9, included angle $40°$:

$\text{Area} = \tfrac{1}{2}(6)(9)\sin 40° = 27 \times 0.6428 \approx 17.4 \text{ cm}^{2}.$

Why? For a right-angled triangle, area = $\tfrac{1}{2}$ base × height. The 'height' from the angle vertex is $a \sin C$ (perpendicular component of side $a$ relative to side $b$). So area = $\tfrac{1}{2} b \cdot (a \sin C) = \tfrac{1}{2}ab\sin C$.

Finding the included angle. If a non-included angle is given:

• Use the angle-sum rule: angles in a triangle sum to $180°$.
• Or use the sine rule to find the included angle indirectly.

Cyclic symmetry. All three labellings give the same area:

$\text{Area} = \tfrac{1}{2} ab \sin C = \tfrac{1}{2} bc \sin A = \tfrac{1}{2} ca \sin B.$

Edexcel tip. Round to 3 sf and INCLUDE UNITS. 'Area $= 17.4$ cm²' scores full marks; '17.4' may lose the final A1.

When it arises. Given two sides and a non-included angle (SSA). The sine rule gives $\sin B = (\ldots)$, but $\sin$ has TWO solutions in $[0°, 180°]$:

• $B_{1} = \sin^{-1}(\ldots)$ (acute, returned by calculator).
• $B_{2} = 180° - B_{1}$ (obtuse).

Which is valid? Check the angle-sum constraint: $A + B + C = 180°$. The angle $B$ must satisfy $B < 180° - (\text{other known angle})$.

Worked example. $XY = 10$, $YZ = 7$, $\angle XZY = 65°$. Find $\angle YXZ$.

• Sine rule: $\sin X / 7 = \sin 65° / 10 \Rightarrow \sin X \approx 0.6344$.
• $X_{1} = \sin^{-1}(0.6344) \approx 39.4°$.
• $X_{2} = 180° - 39.4° = 140.6°$.
• Check: $X_{1} + 65 = 104.4° < 180°$ ✓ valid. $X_{2} + 65 = 205.6° > 180°$ ✗ invalid.
• So $X \approx 39.4°$ only.

When TWO valid solutions exist. If the angle-sum check passes for both, the triangle isn't uniquely determined — there are two possible triangles. Common in exam wording: 'find the two possible values of $\angle X$'.

Geometric intuition. Draw side $YZ$, then a circular arc of radius $XY = 10$ from $Y$. Depending on the geometry, this arc might cross side $XZ$ (extended) at one or two points. Two crossings = two triangles.

$y = \sin x$.

• Period: $360°$ (or $2\pi$ rad).
• Amplitude: 1.
• Range: $[-1, 1]$.
• Key points: $(0, 0)$, $(90°, 1)$, $(180°, 0)$, $(270°, -1)$, $(360°, 0)$.
• Symmetry: odd function — $\sin(-x) = -\sin x$.

$y = \cos x$.

• Period: $360°$.
• Amplitude: 1.
• Range: $[-1, 1]$.
• Key points: $(0, 1)$, $(90°, 0)$, $(180°, -1)$, $(270°, 0)$, $(360°, 1)$.
• Symmetry: even function — $\cos(-x) = \cos x$.

$y = \tan x$.

• Period: $180°$ (twice as often as sine/cosine).
• No amplitude (unbounded).
• Vertical asymptotes at $x = 90°, 270°, \ldots$ (where $\cos x = 0$).
• Key points: $(0, 0)$, $(45°, 1)$, $(180°, 0)$.

Useful identities.

Identity	Use
$\sin^{2}\theta + \cos^{2}\theta = 1$	In booklet
$\sin(180° - \theta) = \sin\theta$	Ambiguous case
$\cos(180° - \theta) = -\cos\theta$	Obtuse cosine
$\tan\theta = \sin\theta / \cos\theta$	Conversion

Edexcel tip. Sketching questions ask you to mark intercepts AND extrema. A common 4-mark question: 'Sketch $y = \sin x$ for $0 \le x \le 360°$, marking the coordinates of all turning points and axis crossings.'

Angle between a line and a plane. The angle between line $L$ and plane $\Pi$ is the angle between $L$ and its PROJECTION onto $\Pi$.

Method.

1. Drop a perpendicular from a point on $L$ down to $\Pi$.
2. The projection is the line from the foot of the perpendicular to where $L$ meets $\Pi$.
3. Use $\tan\theta = \text{height}/\text{base in plane}$.

Worked example: cube of side 2. Find the angle between the diagonal $AG$ and the base $ABCD$ in cube $ABCDEFGH$, where $G$ is the vertex diagonally opposite $A$.

• Project $AG$ onto the base: the projection is $AC$ (the base diagonal).
• $AC = \sqrt{2^{2} + 2^{2}} = 2\sqrt{2}$.
• The vertical leg from $C$ to $G$ is $CG = 2$.
• Required angle: $\tan\theta = 2/(2\sqrt{2}) = 1/\sqrt{2} \Rightarrow \theta \approx 35.3°$.

Angle between two planes. Find a common edge. From a point on this edge, draw two lines — one in each plane, both perpendicular to the edge. The angle between these two lines is the angle between the planes.

Edexcel tip. Always draw a clear 3D sketch labelling all relevant lengths. Use a 2D inset to show the specific right triangle you're using. Mark schemes credit clear identification of the projection.