P1^
7
Exercise(^) 7C
ExAmPlE 7.8 Solve the equation 2sin^2 θ = cos θ + 1 for 0° θ 360°.
SOlUTION
First use the identity sin^2 θ + cos^2 θ = 1 to obtain an equation containing only one
trigonometrical function.
2sin^2 θ = cos θ + 1
⇒ 2(1 − cos^2 θ) = cos θ + 1
⇒ 2 − 2cos^2 θ = cos θ + 1
⇒ 0 = 2cos^2 θ + cos θ − 1
⇒ 0 = (2cos θ − 1)(cos θ + 1)
⇒ 2cos θ − 1 = 0 or cos θ + 1 = 0
⇒ cos θ = 12 or cos θ = −1.
cos θ = 12 ⇒ θ = 60°
or θ = 360° − 60° = 300° (see figure 7.23).
cos θ = − 1 ⇒ θ = 180°.
The values of θ are 60°, 180° or 300°.
ExERCISE 7C 1 (i) Sketch the curve y = sin x for 0° x 360°.
(ii) Solve the equation sin x = 0.5 for 0° x 360°, and illustrate the two roots
on your sketch.
(iii) State the other roots for sin x = 0.5, given that x is no longer restricted to
values between 0° and 360°.
(iv) Write down, without using your calculator, the value of sin 330°.
This is a quadratic
equation in cos θ.
Rearrange it to equal
zero and factorise
it to solve the
equation.
–1
O 60°90° 180° 270°300° 360° θ
1
(^12)
y
y = cos θ
Figure 7.23