Trigonometry

WMA12Topic 2.63 questions

Trigonometric identities and solving trigonometric equations in a given interval.

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Original practice questions written in the style of the exam.

Question 16 marks
Solve, for $0 \leqslant \theta < 360^\circ$ ,

$2\cos^2\theta + 3\sin\theta = 3$
Mark scheme
- M1 Uses $\cos^2\theta = 1 - \sin^2\theta$ (1)
- A1 $2 \sin^{2} θ - 3 \sin θ$ (1)
Question 26 marks
(a) Show that $\dfrac{\sin^2\theta}{1 - \cos\theta} \equiv 1 + \cos\theta$ , for .

+

1

=

0

2\sin^2\theta - 3\sin\theta + 1 = 0

2 sin^{2} θ - 3 sin θ + 1 = 0

M1 Solves the quadratic in

\sin\theta

(2\sin\theta - 1)(\sin\theta - 1) = 0

(1)

\sin\theta = \tfrac{1}{2}

\sin\theta = 1

(1)

\theta = 30^\circ, 150^\circ

(1)

\theta = 90^\circ

(1)

cosθ

\cos\theta \neq 1

(b) Hence solve, for $0 < \theta < 2\pi$ , $\frac{\sin^2\theta}{1 - \cos\theta} = 3\cos\theta$ giving your answers in terms of $\pi$ .

Mark scheme

M1 Replaces $\sin^2\theta$ with $1 - \cos^2\theta$ (1)
M1 Factorises as a difference of two squares: $(1 - \cos\theta)(1 + \cos\theta)$ (1)
A1 Cancels $(1 - \cos\theta)$ to reach $1 + \cos\theta$ with no errors (1)
M1 Uses part (a) to get $1 + \cos\theta = 3\cos\theta$ , so $\cos\theta = \tfrac{1}{2}$ (1)
A1 $\theta = \dfrac{\pi}{3}$ (1)
A1 $\theta = \dfrac{5\pi}{3}$ (1)