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IAL Maths (Edexcel) · P1 · Topic P1.5AS≈ 30 min read

P1.5 Trigonometry: triangles, radians and graphs

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This page covers the sine and cosine rules, the ambiguous case, the area of a triangle, bearings, radian measure, arcs, sectors and segments, and the graphs of sin, cos and tan (Pure Mathematics 1, AS). Solving trigonometric equations and the identities \(\tan x=\dfrac{\sin x}{\cos x}\) and \(\sin^2x+\cos^2x=1\) belong to P2; here you only read solutions off a graph or use symmetry.

What this page covers

8 points
  • The sine rule for sides and for angles, including the ambiguous case where two triangles fit the data
  • The cosine rule for a side or an angle
  • The area of a triangle as \(\tfrac12ab\sin C\)
  • Bearings problems solved with the sine and cosine rules
  • Radian measure and converting between radians and degrees
  • Arc length \(r\theta\), sector area \(\tfrac12r^2\theta\), segment areas and composite shapes
  • The graphs of \(y=\sin x\), \(\cos x\) and \(\tan x\): shape, period and symmetry
  • Simple transformations such as \(3\sin x\), \(\sin\left(x+\tfrac{\pi}{6}\right)\) and \(\sin 2x\)

In one minute

6 points
  • This page gives you the tools for any triangle that is not right-angled, a second unit for measuring angles that makes circle formulae simple, and the three trig graphs you need to sketch from memory.
  • Sine rule: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). Use it when you know a side and the angle opposite it. Finding an angle from its sine gives \(\theta\) and \(180°-\theta\); both can be valid, so check the angle sum.
  • Cosine rule: \(a^2=b^2+c^2-2bc\cos A\). Use it for two sides and the included angle, or for three sides.
  • Area \(=\tfrac12ab\sin C\), where \(C\) is the angle between sides \(a\) and \(b\).
  • Radians: \(\pi\text{ rad}=180°\). With \(\theta\) in radians, arc length \(=r\theta\) and sector area \(=\tfrac12r^2\theta\).
  • Graphs: sin and cos have period \(2\pi\) and range \(-1\) to 1; tan has period \(\pi\) and asymptotes where \(\cos x=0\).

Contents

14 sections
  1. How do you use the sine rule?
  2. When does the sine rule give two triangles?
  3. How do you use the cosine rule and the area formula?
  4. How do you solve a bearings problem?
  5. What is a radian?
  6. How do you find arc lengths, sectors and segments?
  7. What do the graphs of sin, cos and tan look like?
  8. How do simple transformations change a trig graph?
  9. Exam tips for P1.5
  10. Common misconceptions
  11. How this topic connects
  12. Key formulae
  13. Check your understanding
  14. Specification checklist

How do you use the sine rule?

Label the triangle with capital letters for the angles and the matching small letter for the side opposite each angle. Then, in any triangle,

\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\qquad\text{or}\qquad\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\]
  • Use the first form to find a side and the second form to find an angle, so that the unknown sits on top.
  • You need one complete "pair": a side and the angle opposite it.
  • The sine rule is not in the formulae list, so learn it; you will use it constantly.

Why it works: drop a perpendicular from \(C\) to \(AB\). Its height is \(b\sin A\) from one side and \(a\sin B\) from the other, so \(b\sin A=a\sin B\). In a right-angled triangle the rule reduces to sin \(=\) opposite \(\div\) hypotenuse, because \(\sin90°=1\).

Worked example 1 Easy · Find

In triangle \(ABC\), angle \(A=48°\), angle \(B=67°\) and \(BC=11.2\) cm. Find the lengths of \(AC\) and \(AB\), giving each to 3 significant figures.

The third angle: \(C=180°-48°-67°=65°\).

\(AC=b\) is opposite \(B\); the known pair is \(a=11.2\), \(A=48°\). So \(b=\dfrac{11.2\sin67°}{\sin48°}=13.873\ldots\) \(AC=13.9\) cm

\(AB=c\) is opposite \(C\): \(c=\dfrac{11.2\sin65°}{\sin48°}=13.659\ldots\) \(AB=13.7\) cm

Sense check: the longest side, \(b\), is opposite the largest angle, 67°; the shortest, \(a\), is opposite the smallest, 48°.

ABCc = 13.7 cma = 11.2 cmb = 13.9 cm48°67°65°
The triangle of Worked example 1, drawn to scale from the solved lengths. The given pair is \(a=11.2\) cm with \(A=48°\).
Exam tip

Keep full calculator values for lengths you will reuse (store them in memory) and round only the final answers. Rounding \(b\) to 13.9 and using it again can change the third significant figure of a later answer.

When does the sine rule give two triangles?

\(\sin\theta=\sin(180°-\theta)\), so an equation such as \(\sin R=0.82\) has an acute solution and an obtuse one. The calculator only shows the acute one, and you must decide whether the obtuse one also works.

  • The obtuse angle is valid whenever it still leaves room in the triangle: (given angle) \(+\) (obtuse angle) \(<180°\).
  • This can only happen when you are given two sides and an angle that is not between them, and the side opposite the given angle is the shorter of the two given sides.
  • If the side opposite the given angle is the longer one, the obtuse option always fails the angle sum and there is one triangle.
8.0 cm6.06.038°PQR₂R₁55.2°124.8°
Why two triangles fit: the side \(QR=6.0\) cm can swing so that \(R\) lands on the line from \(P\) at two points. Triangle \(PQR_1\) (blue) has \(R=55.2°\); triangle \(PQR_2\) (amber) has \(R=124.8°\).
Worked example 2 Hard · Find

In triangle \(PQR\), \(PQ=8.0\) cm, \(QR=6.0\) cm and angle \(QPR=38°\). (a) Find the two possible sizes of angle \(PRQ\). (b) Find the two possible lengths of \(PR\).

(a) \(QR\) is opposite \(P\) and \(PQ\) is opposite \(R\): \(\sin R=\dfrac{8.0\sin38°}{6.0}=0.8209\).

Both solutions in the range 0° to 180°: \(R=55.2°\) or \(R=180°-55.2°=124.8°\)

Check both fit: \(38°+124.8°=162.8°<180°\) ✓. So the third angle is \(Q=86.8°\) or \(Q=17.2°\).

(b) \(PR\) is opposite \(Q\): \(PR=\dfrac{6.0\sin86.8°}{\sin38°}\) or \(\dfrac{6.0\sin17.2°}{\sin38°}\). \(PR=9.73\) cm or \(2.88\) cm

Common error

Stopping at the calculator's value of \(\sin^{-1}\). When a question says "find the possible values", or the given angle is opposite the shorter given side, always test 180° minus your answer as well.

How do you use the cosine rule and the area formula?

\[a^2=b^2+c^2-2bc\cos A\qquad\qquad\cos A=\frac{b^2+c^2-a^2}{2bc}\]
  • First form: two sides and the angle between them give the third side.
  • Second form: three sides give any angle. A negative cosine means an obtuse angle, and there is never any ambiguity, because \(\cos^{-1}\) gives a unique angle between 0° and 180°.
  • The cosine rule is in the formulae list in its side form; be ready to rearrange it for \(\cos A\).

When \(A=90°\), \(\cos A=0\) and the rule becomes Pythagoras' theorem: the term \(-2bc\cos A\) is the correction for a non-right angle.

Area of a triangle \(=\tfrac12ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the angle between them (the included angle). It is not in the formulae list, so learn it. It is \(\tfrac12\times\) base \(\times\) height with the height written as \(b\sin C\).
You knowUseTo find
a side and its opposite angle, plus one more side or anglesine ruleanother side or angle (watch for the ambiguous case)
two sides and the included anglecosine rulethe third side
three sidescosine rule (cos form)any angle; find the largest (opposite the longest side) first
two sides and the included angle\(\tfrac12ab\sin C\)the area
Worked example 3 Medium · Find

A triangular field has sides of length 85 m, 120 m and 150 m. (a) Find the largest angle of the field. (b) Find the area of the field.

(a) The largest angle is opposite the longest side, 150 m: \(\cos\theta=\dfrac{85^2+120^2-150^2}{2\times85\times120}=\dfrac{-875}{20\,400}=-0.04289\). \(\theta=92.5°\)

(b) Use \(\tfrac12ab\sin C\) with the two sides that enclose \(\theta\): Area \(=\tfrac12\times85\times120\times\sin92.5°=5095.3\ldots\) 5100 m² (3 s.f.)

The negative cosine signals an obtuse angle; the calculator handles this automatically with \(\cos^{-1}\).

85 m150 m120 mθarea ≈ 5100 m²
The field drawn to scale. The largest angle is always opposite the longest side, so find that one first: once it is known, the other two must be acute.

How do you solve a bearings problem?

  • A bearing is measured clockwise from north and written with three figures: 072°, 195°.
  • Draw a north line at every point you measure from.
  • The bearing of \(A\) from \(B\) and the bearing of \(B\) from \(A\) differ by exactly 180° (the back bearing).
  • North lines are parallel, so use alternate and co-interior angles to find the angles inside the triangle, then use the sine or cosine rule.
Worked example 4 Hard · Show that, Find

A yacht leaves harbour \(H\) and sails 14 km on a bearing of 072° to a buoy \(B\). It then sails 9 km on a bearing of 195° to a point \(C\). (a) Show that angle \(HBC=57°\). (b) Find the distance \(HC\). (c) Find the bearing of \(C\) from \(H\). (d) Write down the bearing on which the yacht must sail to return directly from \(C\) to \(H\).

(a) The bearing of \(H\) from \(B\) is the back bearing: \(72°+180°=252°\), so angle \(HBC=252°-195°=57°\) as required.

(b) Two sides and the included angle, so use the cosine rule: \(HC^2=14^2+9^2-2\times14\times9\times\cos57°=139.75\), so \(HC=11.822\ldots\) \(HC=11.8\) km

(c) Angle \(BHC\) by the sine rule (it is opposite the 9 km side, so it is acute): \(\sin BHC=\dfrac{9\sin57°}{11.822}\), so angle \(BHC=39.7°\). Bearing of \(C\) from \(H=072°+39.7°=\) 111.7°

(d) Back bearing: \(111.7°+180°=\) 291.7°

NN14 km9 km11.8 km72°195°57°HBC
Worked example 4 to scale. The angle inside the triangle at \(B\) (amber) is the back bearing 252° minus 195°. The dashed line \(HC\) is 11.8 km on a bearing of 111.7° from \(H\).
Common error

Using the bearing itself as an angle in the triangle. A bearing is an angle from north, not between two legs of the journey: always draw the north lines and work out the angle inside the triangle first.

What is a radian?

One radian is the angle at the centre of a circle subtended by an arc whose length equals the radius.
1 radrrarc = rO
An arc of length \(r\) subtends an angle of 1 radian at the centre, whatever the size of the circle. That is why radians make arc and sector formulae so simple.
  • A full turn is an arc of \(2\pi r\), which is \(2\pi\) radii, so \(2\pi\text{ rad}=360°\) and \(\pi\text{ rad}=180°\).
  • Degrees → radians: multiply by \(\dfrac{\pi}{180}\). Radians → degrees: multiply by \(\dfrac{180}{\pi}\).
  • \(1\text{ rad}=\dfrac{180°}{\pi}\approx57.3°\).
  • "rad" is often left out: an angle written without the ° symbol is in radians.
  • Set your calculator to radian mode whenever the angles are in radians, including inside the sine and cosine rules.
Degrees30°45°60°90°120°135°180°270°360°
Radians\(\tfrac{\pi}{6}\)\(\tfrac{\pi}{4}\)\(\tfrac{\pi}{3}\)\(\tfrac{\pi}{2}\)\(\tfrac{2\pi}{3}\)\(\tfrac{3\pi}{4}\)\(\pi\)\(\tfrac{3\pi}{2}\)\(2\pi\)

How do you find arc lengths, sectors and segments?

With \(\theta\) in radians, \(r\) the radius, \(s\) the arc length and \(A\) the sector area:

\[s=r\theta\qquad\qquad A=\tfrac12r^2\theta\]

Neither is in the formulae list, so learn both. They come from fractions of a circle: the sector is \(\dfrac{\theta}{2\pi}\) of the whole, so \(s=\dfrac{\theta}{2\pi}\times2\pi r\) and \(A=\dfrac{\theta}{2\pi}\times\pi r^2\). In degrees you would need extra factors of \(\dfrac{\pi}{180}\).

Worked example 5 Easy · Write, Find

(a) Write 135° in radians as an exact multiple of \(\pi\), and 2.4 rad in degrees. (b) A sector \(OAB\) has radius 9 cm and angle \(AOB=1.2\) rad. Find its arc length, perimeter and area.

(a) \(135\times\dfrac{\pi}{180}=\) \(\dfrac{3\pi}{4}\) and \(2.4\times\dfrac{180}{\pi}=137.51\ldots\) 137.5°

(b) Arc length \(s=9\times1.2=\) 10.8 cm. The perimeter includes both radii: \(9+9+10.8=\) 28.8 cm. Area \(=\tfrac12\times9^2\times1.2=\) 48.6 cm²

1.2 rad9 cm9 cmarc 10.8 cm48.6 cm²OAB
The sector of Worked example 5. Its perimeter is two radii plus the arc, which is why 28.8 cm is much more than the arc alone.

Segments

  • A segment is the region between a chord and its arc.
  • Segment \(=\) sector \(-\) triangle: \(\tfrac12r^2\theta-\tfrac12r^2\sin\theta=\tfrac12r^2(\theta-\sin\theta)\).
  • The triangle is isosceles with two sides \(r\) and the included angle \(\theta\), so its area is \(\tfrac12r^2\sin\theta\) (with sin of an angle in radians).
  • The perimeter of a segment is the chord plus the arc.
Worked example 6 Medium · Show that, Find

A chord \(AB\) of length 10 cm is drawn in a circle with centre \(O\) and radius 7 cm. Angle \(AOB=\theta\) radians. (a) Show that \(\theta=1.591\) to 4 significant figures. (b) Find the area of the minor segment cut off by \(AB\). (c) Find the perimeter of this segment.

(a) Cosine rule in triangle \(OAB\): \(\cos\theta=\dfrac{7^2+7^2-10^2}{2\times7\times7}=\dfrac{-2}{98}=-\dfrac{1}{49}\). So \(\theta=\cos^{-1}\left(-\tfrac{1}{49}\right)=1.59121\ldots=1.591\) (4 s.f.) as required.

(b) Sector minus triangle, calculator in radians: \(\tfrac12\times7^2\times1.5912-\tfrac12\times7^2\times\sin1.5912=38.98-24.49=14.490\ldots\) 14.5 cm²

(c) Chord plus arc: \(10+7\times1.5912=21.138\ldots\) 21.1 cm

A quicker route to (a): half the chord gives \(\sin\dfrac{\theta}{2}=\dfrac57\), so \(\theta=2\sin^{-1}\dfrac57\), the same value.

710 cm7θsegmentOAB
Worked example 6 to scale: radius 7 cm, chord 10 cm, \(\theta=1.591\) rad. The sector is the triangle (blue) plus the segment (amber).
Worked example 7 Hard · Find

A flower bed is made from a sector \(OAB\) of a circle, centre \(O\), radius 6 m, with angle \(AOB=1.1\) rad, joined along \(OB\) to a triangle \(OBC\) in which \(OC=8\) m and angle \(BOC=0.9\) rad. The two pieces do not overlap. (a) Find the length \(BC\). (b) Find the area of the flower bed. (c) Find the perimeter of the flower bed.

(a) Cosine rule with the angle in radians (calculator in radian mode): \(BC^2=6^2+8^2-2\times6\times8\times\cos0.9=40.325\), so \(BC=6.350\ldots\) \(BC=6.35\) m

(b) Sector \(+\) triangle: \(\tfrac12\times6^2\times1.1+\tfrac12\times6\times8\times\sin0.9=19.8+18.80=\) 38.6 m²

(c) Boundary \(=OA+\) arc \(AB+BC+CO\) (\(OB\) is inside the bed): \(6+6\times1.1+6.350+8=26.95\ldots\) 27.0 m

In degree mode, \(\cos0.9\) would be the cosine of 0.9°, and \(BC\) would come out as about 2.00 m: a quick way to spot a mode error.

66.35 m8 m1.10.96 mOABC
The flower bed: a sector (teal) joined along \(OB\) to a triangle (blue). \(OB\) lies inside the shape, so it is not part of the perimeter.
Exam tip

For a composite shape, sketch it and mark which lines are on the outside. Radii and chords that lie inside the shape are not part of the perimeter, but they are often the key to splitting the area into sectors and triangles.

What do the graphs of sin, cos and tan look like?

π/2π3π/22π−1−0.50.51y = sin xy = cos x
\(y=\sin x\) and \(y=\cos x\) for \(0\le x\le2\pi\). The cosine curve is the sine curve shifted \(\tfrac{\pi}{2}\) to the left.
π/2π3π/22π−4−224x = π/2x = 3π/2y = tan x
\(y=\tan x\) for \(0\le x\le2\pi\). It rises through every interval between its asymptotes at \(x=\tfrac{\pi}{2}\) and \(x=\tfrac{3\pi}{2}\), and repeats every \(\pi\).
\(y=\sin x\)\(y=\cos x\)\(y=\tan x\)
Period\(2\pi\) (360°)\(2\pi\) (360°)\(\pi\) (180°)
Range\(-1\le y\le1\)\(-1\le y\le1\)all real numbers
Zeros\(0, \pi, 2\pi, \ldots\)\(\tfrac{\pi}{2}, \tfrac{3\pi}{2}, \ldots\)\(0, \pi, 2\pi, \ldots\)
Maximum / minimum1 at \(\tfrac{\pi}{2}\); \(-1\) at \(\tfrac{3\pi}{2}\)1 at 0, \(2\pi\); \(-1\) at \(\pi\)none
Asymptotesnonenone\(x=\tfrac{\pi}{2}, \tfrac{3\pi}{2}, \ldots\)
Symmetry\(\sin(\pi-x)=\sin x\); odd: \(\sin(-x)=-\sin x\)\(\cos(2\pi-x)=\cos x\); even: \(\cos(-x)=\cos x\)\(\tan(\pi+x)=\tan x\); odd

Exact values to know

0\(\tfrac{\pi}{6}\) (30°)\(\tfrac{\pi}{4}\) (45°)\(\tfrac{\pi}{3}\) (60°)\(\tfrac{\pi}{2}\) (90°)
sin0\(\tfrac12\)\(\tfrac{\sqrt2}{2}\)\(\tfrac{\sqrt3}{2}\)1
cos1\(\tfrac{\sqrt3}{2}\)\(\tfrac{\sqrt2}{2}\)\(\tfrac12\)0
tan0\(\tfrac{\sqrt3}{3}\)1\(\sqrt3\)undefined

Using symmetry: from the graph, \(\sin\tfrac{5\pi}{6}=\sin\left(\pi-\tfrac{\pi}{6}\right)=\tfrac12\) and \(\cos\tfrac{4\pi}{3}=-\cos\tfrac{\pi}{3}=-\tfrac12\). The curves repeat, so \(\sin(x+2\pi)=\sin x\). The number of solutions of an equation like \(\sin x=k\) in an interval is the number of times the line \(y=k\) crosses the curve there.

Common error

Drawing \(\tan x\) with a maximum and minimum like \(\sin x\), or letting it cross its asymptotes. The tan curve rises through every interval between asymptotes and repeats every \(\pi\), not every \(2\pi\).

How do simple transformations change a trig graph?

EquationTransformation of \(y=\sin x\)Effect
\(y=3\sin x\)stretch parallel to the \(y\)-axis, scale factor 3range \(-3\le y\le3\); zeros and period unchanged
\(y=\sin x+2\)translation 2 uprange \(1\le y\le3\)
\(y=\sin\left(x+\tfrac{\pi}{6}\right)\)translation \(\tfrac{\pi}{6}\) to the leftevery point moves left by \(\tfrac{\pi}{6}\); period unchanged
\(y=\sin2x\)stretch parallel to the \(x\)-axis, scale factor \(\tfrac12\)period \(2\pi\div2=\pi\); range unchanged
\(y=-\sin x\)reflection in the \(x\)-axismaxima become minima
  • Changes outside the function act on \(y\) as you expect; changes inside act on \(x\) the "opposite" way: \(+\tfrac{\pi}{6}\) moves left, \(\times2\) squashes.
  • For \(y=\sin bx\) and \(y=\cos bx\) the period is \(\dfrac{2\pi}{b}\); for \(y=\tan bx\) it is \(\dfrac{\pi}{b}\).
  • To find new key points, solve "inside \(=\) old key value": the maximum of \(\sin\left(x+\tfrac{\pi}{6}\right)\) is where \(x+\tfrac{\pi}{6}=\tfrac{\pi}{2}\), so \(x=\tfrac{\pi}{3}\).
π/2π3π/22π−3−2−1123y = 3 sin xy = sin x(π/2, 3)(3π/2, −3)
\(y=3\sin x\) over \(y=\sin x\) (dashed): only the height changes. The zeros at \(0\), \(\pi\) and \(2\pi\) stay where they were.
π/3π/2π4π/33π/22π−1−0.50.51y = sin(x + π/6)y = sin x(π/3, 1)(0, ½)
\(y=\sin\left(x+\tfrac{\pi}{6}\right)\) over \(y=\sin x\) (dashed): only the position changes. The maximum moves from \(\tfrac{\pi}{2}\) to \(\tfrac{\pi}{3}\), and the curve now meets the \(y\)-axis at \(\left(0, \tfrac12\right)\).
Worked example 8 Hard · Describe, Find, State

The curve \(C\) has equation \(y=\cos\left(x+\tfrac{\pi}{3}\right)\), \(0\le x\le2\pi\). (a) Describe fully the transformation that maps \(y=\cos x\) onto \(C\). (b) Find the exact coordinates of the points where \(C\) meets the coordinate axes. (c) Find the coordinates of the maximum and minimum points of \(C\). (d) State the number of solutions of \(\cos\left(x+\tfrac{\pi}{3}\right)=-0.4\) in the interval \(0\le x\le2\pi\).

(a) A translation by \(\tfrac{\pi}{3}\) in the negative \(x\)-direction, vector \(\begin{pmatrix}-\tfrac{\pi}{3}\\0\end{pmatrix}\)

(b) \(y\)-axis: put \(x=0\), so \(y=\cos\tfrac{\pi}{3}=\tfrac12\), giving \(\left(0, \tfrac12\right)\). \(x\)-axis: cos \(=0\) when the inside is \(\tfrac{\pi}{2}\) or \(\tfrac{3\pi}{2}\), so \(x+\tfrac{\pi}{3}=\tfrac{\pi}{2}\) or \(\tfrac{3\pi}{2}\). \(\left(0, \tfrac12\right)\), \(\left(\tfrac{\pi}{6}, 0\right)\) and \(\left(\tfrac{7\pi}{6}, 0\right)\)

(c) The inside runs from \(\tfrac{\pi}{3}\) to \(\tfrac{7\pi}{3}\). cos \(=-1\) when the inside is \(\pi\); cos \(=1\) when the inside is \(2\pi\). Minimum \(\left(\tfrac{2\pi}{3}, -1\right)\), maximum \(\left(\tfrac{5\pi}{3}, 1\right)\)

(d) The line \(y=-0.4\) cuts the curve on either side of the minimum: 2 solutions

π/62π/37π/65π/32π−1−0.50.51y = −0.4(0, ½)(π/6, 0)(7π/6, 0)(2π/3, −1)(5π/3, 1)y = cos(x + π/3)
Worked example 8 plotted from \(y=\cos\left(x+\tfrac{\pi}{3}\right)\). The amber points are the 2 intersections with \(y=-0.4\), at \(x\approx0.935\) and \(3.254\).
Exam tip

When you sketch a transformed trig graph, label every intercept with the axis and every maximum and minimum with exact coordinates, and make the curve start and stop exactly at the ends of the stated interval.

Exam tips for P1.5

Exam tip 1 — check the mode before you start

Degrees for triangles given in degrees and for bearings; radians for arcs, sectors and graphs on \(0\le x\le2\pi\). Glance at the question for a ° symbol or a \(\pi\) and set the calculator first.

Exam tip 2 — choose the rule from what you know

Before writing anything, mark the given information on a sketch. A complete opposite pair means the sine rule; two sides with the angle between them, or three sides, means the cosine rule.

Exam tip 3 — exact means exact

"Give your answer in the form \(a\pi+b\sqrt3\)" or "exact" means keep \(\pi\) and surds: use the exact values \(\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}\) and so on, and do not switch to decimals.

Exam tip 4 — show-that answers need more accuracy

To show that \(\theta=1.591\), write a value to more figures (1.5912…) before rounding. Simply quoting the target value does not show anything.

Exam tip 5 — read the command word

"Sketch" needs the correct shape with key points labelled, not an accurate plot; "write down" or "state" means little or no working is needed; "find" and "calculate" need visible working; "show that" needs every step.

Common misconceptions

MisconceptionWhy it is wrongCorrect idea
The sine rule only ever gives one angle.\(\sin\theta=\sin(180°-\theta)\), and the calculator shows only the acute value.Test the obtuse value \(180°-\theta\) against the angle sum.
\(\tfrac12ab\sin C\) works with any angle.The formula needs the angle between sides \(a\) and \(b\).Use the included angle, or find it first.
A bearing is the angle between two paths.Bearings are measured clockwise from north.Draw north lines, then find the interior angles.
\(s=r\theta\) works with \(\theta\) in degrees.The formula comes from \(2\pi r\) for \(2\pi\) radians.Convert to radians first (or use \(\tfrac{\theta}{360}\times2\pi r\)).
\(\sin\left(x+\tfrac{\pi}{6}\right)\) is \(\sin x\) moved \(\tfrac{\pi}{6}\) to the right.The new curve reaches each value when \(x+\tfrac{\pi}{6}\) equals the old \(x\), that is \(\tfrac{\pi}{6}\) earlier.\(+\) inside the bracket moves the graph left.
\(\sin2x\) has twice the period of \(\sin x\).The inside reaches \(2\pi\) when \(x=\pi\).Period \(\tfrac{2\pi}{2}=\pi\): the curve is squashed horizontally.

How this topic connects

P2 Pure Mathematics 2

The graphs and symmetries here are what you use to find every solution of a trigonometric equation in an interval, and the identities \(\tan x=\dfrac{\sin x}{\cos x}\) and \(\sin^2x+\cos^2x=1\) build on the same unit-circle ideas.

P1.3 Transformations, P3 and P4

The four transformation rules of P1.3 are exactly the ones used on trig graphs here. Radians are essential in calculus: the derivative of \(\sin x\) is \(\cos x\) only when \(x\) is in radians, and combined transformations such as \(R\sin(x+\alpha)\) extend the last section.

M1 Mechanics and IAL Physics

Resolving forces, finding resultants and velocity triangles use the sine and cosine rules and bearings-style directions. Angular displacement and angular speed in circular motion are measured in radians, and oscillations follow sine and cosine graphs.

Key formulae

FormulaUseIn the formulae list?
\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)sine rule: sides and angles in any triangleNo – learn it
\(a^2=b^2+c^2-2bc\cos A\)cosine rule: third side, or an angle from three sidesYes
Area \(=\tfrac12ab\sin C\)area of a triangle from two sides and the included angleNo – learn it
\(\pi\text{ rad}=180°\)converting between radians and degreesNo – learn it
\(s=r\theta\)arc length, \(\theta\) in radiansNo – learn it
\(A=\tfrac12r^2\theta\)sector area, \(\theta\) in radiansNo – learn it
\(\tfrac12r^2(\theta-\sin\theta)\)segment areaNo – derive it
period \(\dfrac{2\pi}{b}\) (sin, cos); \(\dfrac{\pi}{b}\) (tan)period of \(\sin bx\), \(\cos bx\), \(\tan bx\)No – derive it

Check your understanding

1. Write 225° in radians as an exact multiple of \(\pi\), and convert 0.6 rad to degrees.

\(225\times\dfrac{\pi}{180}=\) \(\dfrac{5\pi}{4}\). \(0.6\times\dfrac{180}{\pi}=\) 34.4°.

2. Find the area of a triangle with sides 7 cm and 9 cm and an angle of 110° between them.

\(\tfrac12\times7\times9\times\sin110°=29.60\ldots=\) 29.6 cm² (3 s.f.).

3. A triangle has sides 5, 6 and 8 units. Find its largest angle.

Opposite the side of 8: \(\cos\theta=\dfrac{25+36-64}{60}=-0.05\), so \(\theta=92.9°\).

4. A sector of radius 5 cm has area 15 cm². Find the angle of the sector and its arc length.

\(\tfrac12\times25\times\theta=15\) gives \(\theta=1.2\) rad; arc \(=5\times1.2=\) 6 cm.

5. State the period of \(y=\sin3x\) and of \(y=\tan2x\), and the exact value of \(\sin\tfrac{5\pi}{6}\).

\(\tfrac{2\pi}{3}\) and \(\tfrac{\pi}{2}\). By symmetry, \(\sin\tfrac{5\pi}{6}=\sin\left(\pi-\tfrac{\pi}{6}\right)=\sin\tfrac{\pi}{6}=\) \(\tfrac12\).

6. In triangle \(XYZ\), \(XY=10\) cm, \(YZ=12\) cm and angle \(X=50°\). Explain why there is only one possible triangle.

The side opposite the given angle (\(YZ=12\)) is the longer of the two given sides. \(\sin Z=\dfrac{10\sin50°}{12}=0.638\), so \(Z=39.7°\) or \(140.3°\); but \(50°+140.3°>180°\), so only \(Z=39.7°\) works and there is one triangle.

Specification checklist

Tick every box before the exam. The wording is ours; the numbers refer to the P1 specification sections.

  • (P1 3.1) I can use the sine rule to find a side or an angle, and I can find both triangles when the data are ambiguous.
  • (P1 3.1) I can use the cosine rule to find a side from two sides and the included angle, or an angle from three sides.
  • (P1 3.1) I can find the area of any triangle using \(\tfrac12ab\sin C\).
  • (P1 3.1) I can turn a bearings description into a triangle and solve it with the sine and cosine rules.
  • (P1 3.2) I can convert between degrees and radians, and I know the radian values of common angles exactly.
  • (P1 3.2) I can use arc length \(r\theta\) and sector area \(\tfrac12r^2\theta\), and find the areas and perimeters of segments and composite shapes.
  • (P1 3.3) I can sketch \(y=\sin x\), \(y=\cos x\) and \(y=\tan x\), and use their periods and symmetries.
  • (P1 3.3) I can sketch simple transformations such as \(y=3\sin x\), \(y=\sin\left(x+\tfrac{\pi}{6}\right)\) and \(y=\sin2x\), labelling intercepts, maxima and minima.