Free Pearson Edexcel International A-level Mathematics · P1

Triangles, radians and trigonometric graphs

What you’ll learn

  • Solve triangle problems with sine and cosine rules.
  • Use radian measure for arcs and sectors.
  • Recognise trigonometric periods and exact values.

Understand the idea

Choose the cosine rule when two sides and their included angle are known, or when all three sides are known. The sine rule links a side to its opposite angle. With two sides and a non-included angle, a second triangle may exist because supplementary angles have the same sine.

One radian subtends an arc equal in length to the radius. This makes arc and sector calculations compact, but their standard formulae require radians. Convert degrees by multiplying by π/180\pi/180. A segment area is the sector area minus the triangle area.

Sine and cosine repeat every 2π2\pi radians, while tangent repeats every π\pi. The multiplier inside the angle changes the period: sin⁡(kx)\sin(kx) completes a cycle in 2π/∣k∣2\pi/|k|. The multiplier outside changes amplitude, and a vertical shift changes the midline.

Key formulae

asin⁡A=bsin⁡B=csin⁡C\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}

c2=a2+b2−2abcos⁡Cc^2=a^2+b^2-2ab\cos C

triangle area=12absin⁡C\text{triangle area}=\frac12ab\sin C

s=rθ,sector area=12r2θ(θ in radians)s=r\theta,\qquad \text{sector area}=\frac12r^2\theta\quad(\theta\text{ in radians})

One full sine cycle spans 2π radians

Conditions. Exact sine function over 0 to 2π radians. Tick values are rounded numerically; π ≈ 3.142. The amplitude is 1 and the period is 2π.

Angle / rad; sin(angle) / dimensionless.

Zeros occur at 0, π and 2π. The maximum is 1 at π/2 and minimum −1 at 3π/2. Degree input would require conversion before evaluating this radian model.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

A circle has radius 88 cm. Find the arc length and sector area for an angle of 75∘75^\circ.

  1. Convert to radians: 75π/180=5π/1275\pi/180=5\pi/12.
  2. Arc length is 8(5π/12)=10π/38(5\pi/12)=10\pi/3.
  3. Sector area is 12(82)(5π/12)=40π/3\frac12(8^2)(5\pi/12)=40\pi/3.

Answer: Arc length 10π3\frac{10\pi}{3} cm; sector area 40π3\frac{40\pi}{3} cm2^2.

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