E1 - Trigonometric ratios and triangle rules | AQA A-Level Mathematics

Give trigonometric ratios a meaning for every rotation, choose a rule from the information in a triangle, and calculate its area. You will also discover why two sides and one angle can describe two different triangles.

Ratios for every angle

Can an angle have positive sine but negative cosine, and can the same triangle measurements describe two triangles? A radius ending at $(-3,4)$ answers the first question: its vertical coordinate is positive and its horizontal coordinate is negative. We will answer the second by looking carefully at the sine rule.

You already know

In a right triangle, sine is opposite divided by hypotenuse, cosine is adjacent divided by hypotenuse, and tangent is opposite divided by adjacent. Pythagoras gives $r^2=x^2+y^2$ for perpendicular sides $x,y$ and hypotenuse $r$ . These results are not in the formula booklet: learn them.

A right triangle only describes acute angles. A rotating radius extends the same ratios: start on the positive horizontal axis, turn anticlockwise for a positive angle and clockwise for a negative angle. Extra complete turns are allowed.

Trigonometric ratios for a rotation

If a terminal radius passes through $(x,y)\ne(0,0)$ , let $r=\sqrt{x^2+y^2}>0$ . Then

\sin\theta=\frac{y}{r},\qquad \cos\theta=\frac{x}{r},\qquad \tan\theta=\frac{y}{x}\quad(x\ne0)

On the unit circle, $r=1$ , so the terminal point is $(\cos\theta,\sin\theta)$ . The origin cannot define an angle this way because its radius is zero.

These definitions are not supplied as formulae in the booklet: learn them. Sine and cosine exist for every real angle and lie in $[-1,1]$ . Tangent is undefined at $90^\circ+180^\circ k$ , for integer $k$ , because the horizontal coordinate is zero. An undefined value is neither zero nor infinity.

Terminal quadrant	Sine	Cosine	Tangent
I: $0^\circ<\theta<90^\circ$	positive	positive	positive
II: $90^\circ<\theta<180^\circ$	positive	negative	negative
III: $180^\circ<\theta<270^\circ$	negative	negative	positive
IV: $270^\circ<\theta<360^\circ$	negative	positive	negative

At an axis, use coordinates directly: $180^\circ$ ends at $(-1,0)$ , giving sine $0$ , cosine $-1$ and tangent $0$ . A full turn returns to the same point. A half-turn changes both coordinate signs, so tangent already repeats after $180^\circ$ . These symmetry facts are not in the formula booklet: learn them.

Worked example 1: signed coordinates

Question. A terminal radius passes through $(-3,4)$ . Find its sine, cosine and tangent exactly.

Plan. Find the positive radius, then use the signed coordinate ratios.

\begin{aligned} r&=\sqrt{(-3)^2+4^2}=5 && \text{Pythagoras} \\ \sin\theta&=\frac45 && \text{vertical over radius} \\ \cos\theta&=-\frac35 && \text{horizontal over radius} \\ \tan\theta&=-\frac43 && \text{vertical over horizontal} \end{aligned}

Why this step? The radius is a length and stays positive; the coordinates carry the signs.

Answer. $\sin\theta=4/5$ , $\cos\theta=-3/5$ , $\tan\theta=-4/3$ .

Check. The point is in quadrant II, where sine is positive and cosine and tangent are negative.

Change only the sign of the vertical coordinate. The radius stays the same, but two of the ratios change sign.

Worked example 2: reflecting the terminal point

Question. A terminal radius passes through $(-3,-4)$ . Find its sine, cosine and tangent exactly.

Plan. Use the same radius and keep the negative signs in both coordinates.

\begin{aligned} r&=\sqrt{(-3)^2+(-4)^2}=5 && \text{squares are unchanged} \\ \sin\theta&=-\frac45 && \text{vertical over radius} \\ \cos\theta&=-\frac35 && \text{horizontal over radius} \\ \tan\theta&=\frac43 && \text{two negatives divide} \end{aligned}

Why this step? Tangent uses two signed coordinates, so it is positive in quadrant III.

Answer. $\sin\theta=-4/5$ , $\cos\theta=-3/5$ , $\tan\theta=4/3$ .

Check. Both sine and cosine are negative, matching the terminal point.

Notice that tangent can have magnitude greater than one: its denominator is a coordinate, not the radius.

Watch out

Wrong: $r=-5$ , so $\sin\theta=4/5$ at $(-3,-4)$ . A radius is a positive length. Use $r=5$ and retain the negative vertical coordinate: $\sin\theta=-4/5$ .

For your turn, change the coordinate magnitudes while keeping quadrant III.

Choosing a triangle rule

A triangle with two sides $7\,\mathrm{cm}$ and $10\,\mathrm{cm}$ and their included angle $60^\circ$ has no known opposite side–angle pair. That makes the cosine rule the natural starting point.

Write $a,b,c$ opposite $A,B,C$ . Triangle angles are strictly between $0^\circ$ and $180^\circ$ and sum to $180^\circ$ . The largest side must be less than the sum of the other two: equality makes a straight line.

Sine and cosine rules

For a non-degenerate triangle,

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

c^2=a^2+b^2-2ab\cos C,\qquad \cos C=\frac{a^2+b^2-c^2}{2ab}

The corresponding cyclic versions use the other opposite pairs.

Not in the formula booklet: learn it.

The sine rule comes from expressing the same height in two ways: $b\sin A=a\sin B$ , then dividing by $\sin A\sin B$ . The cosine rule comes from the squared distance between $(a,0)$ and $(b\cos C,b\sin C)$ . Expansion gives $a^2+b^2-2ab\cos C$ , using the recall identity $\sin^2C+\cos^2C=1$ , which is not in the formula booklet.

Information	Useful starting rule
Two sides and their included angle	Cosine rule for the third side
Three sides	Cosine rule for an angle
A complete opposite pair and another side or angle	Sine rule; check ambiguity if finding an angle

At $C=90^\circ$ , the cosine term vanishes, leaving Pythagoras. A negative cosine correctly makes the side opposite an obtuse angle longer.

Worked example 3: an included acute angle

Question. Sides $7\,\mathrm{cm}$ and $10\,\mathrm{cm}$ meet at $60^\circ$ . Find the opposite side exactly.

Plan. Use the cosine rule and take the positive square root.

\begin{aligned} c^2&=7^2+10^2-2(7)(10)\cos60^\circ && \text{included angle} \\ &=149-70=79 && \cos60^\circ=\frac12 \\ c&=\sqrt{79}\,\mathrm{cm} && \text{positive length} \end{aligned}

Why this step? The known angle lies between the two known sides, so no opposite pair is needed.

Answer. $c=\sqrt{79}\,\mathrm{cm}$ .

Check. $3<\sqrt{79}<17$ , so the triangle inequalities hold.

Change only the included angle from $60^\circ$ to $120^\circ$ . The negative cosine now increases the squared side.

Worked example 4: an included obtuse angle

Question. Sides $7\,\mathrm{cm}$ and $10\,\mathrm{cm}$ meet at $120^\circ$ . Find the opposite side exactly.

Plan. Use the same cosine rule with the new angle.

\begin{aligned} c^2&=149-140\cos120^\circ && \text{cosine rule} \\ &=149+70=219 && \cos120^\circ=-\frac12 \\ c&=\sqrt{219}\,\mathrm{cm} && \text{positive root} \end{aligned}

Why this step? Subtracting a negative term gives a longer opposite side, as the wider angle suggests.

Answer. $c=\sqrt{219}\,\mathrm{cm}$ .

Check. $10<\sqrt{219}<17$ ; the side opposite the obtuse angle is the longest.

When a complete opposite pair is known, switch to the sine rule. The next pair keeps the angles fixed and changes only the scale of the triangle.

Worked example 5: a known opposite pair

Question. In triangle $ABC$ , $A=35^\circ$ , $B=65^\circ$ and $a=8\,\mathrm{cm}$ . Find $b$ to three significant figures.

Plan. Match $a$ with $A$ , then rearrange the sine rule for $b$ .

\begin{aligned} \frac{b}{\sin65^\circ}&=\frac8{\sin35^\circ} && \text{opposite pairs} \\ b&=\frac{8\sin65^\circ}{\sin35^\circ} && \text{rearrange} \\ &=12.6407\ldots\,\mathrm{cm} && \text{degree mode} \end{aligned}

Why this step? Both angles are known, so finding a side introduces no inverse-sine ambiguity.

Answer. $b=12.6\,\mathrm{cm}$ , to three significant figures.

Check. $B>A$ and $b>a$ , as required.

Double only the given side. Similar triangles predict that the answer doubles too.

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E1 - Trigonometric ratios and triangle rules

Ratios for every angle

Choosing a triangle rule

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