Cambridge International AS and A Level Mathematics Pure Mathematics 1

Trigonometry

P1^

7

●?^ How can you find further roots of the equation 3tan θ^ =^ −1, outside the range

−180°  θ  180°?

ExAmPlE 7.7  Find values of θ in the interval 0º  θ  360 º for which tan^2 θ − tan θ = 2.

SOlUTION

First rearrange the equation.

tan^2 θ − tan θ = 2

⇒ tan^2 θ − tan θ − 2 = 0

⇒ (tan θ − 2)(tan θ + 1) = 0

⇒ tan θ = 2 or tan θ = −1.

tan θ = 2 ⇒ θ = 63.4º (calculator)

or θ = 63.4º + 180 º (see figure 7.22)

= 243.4º.

tan θ = − 1 ⇒    θ = − 45 º (calculator).

This is not in the range 0°  θ  360° so figure 7.22 is used to give

θ = −45° + 180° = 135°

or θ = −45° + 360° = 315°.

The values of θ are 63.4°, 135°, 243.4°, 315°.

This is a quadratic equation

like x^2 – x – 2 = 0.

θ

tan θ

90° 270°

–1

O 180°

2

360°

Figure 7.22