Edexcel A level Maths (9MA0) · Pure mathematics › Trigonometry
Practise sec, cosec, cot and inverse trig functions. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Find the exact value of \(\sec\left(\frac{3\pi}{4}\right)\).[2]
(b) Find the exact value of \(\operatorname{cosec}\left(\frac{5\pi}{6}\right)\).[2]
(c) Find the exact value of \(\cot\left(\frac{7\pi}{6}\right)\).[2]
Show the answer and mark scheme
(a)Answer: \(-\sqrt{2}\)
M1 for \(\sec\left(\frac{3\pi}{4}\right) = \frac{1}{\cos\left(\frac{3\pi}{4}\right)}\) with \(\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\)
A1 for \(-\sqrt{2}\) oe
Worked solution: \(\sec\left(\frac{3\pi}{4}\right) = \frac{1}{\cos\left(\frac{3\pi}{4}\right)} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}\)
(b)Answer: \(2\)
M1 for \(\operatorname{cosec}\left(\frac{5\pi}{6}\right) = \frac{1}{\sin\left(\frac{5\pi}{6}\right)}\) with \(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)
A1 for \(2\) oe
Worked solution: \(\operatorname{cosec}\left(\frac{5\pi}{6}\right) = \frac{1}{\sin\left(\frac{5\pi}{6}\right)} = \frac{1}{\frac{1}{2}} = 2\)
(c)Answer: \(\sqrt{3}\)
M1 for \(\cot\left(\frac{7\pi}{6}\right) = \frac{1}{\tan\left(\frac{7\pi}{6}\right)}\) with \(\tan\left(\frac{7\pi}{6}\right) = \frac{\sqrt{3}}{3}\)
A1 for \(\sqrt{3}\) oe
Worked solution: \(\cot\left(\frac{7\pi}{6}\right) = \frac{1}{\tan\left(\frac{7\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}\)
Question 2Medium4 marks
Solve, for \(0 \le x \lt \pi\), the equation \[\operatorname{cosec}\left(2x + \frac{\pi}{3}\right) = 4.\] Give your answers to 3 decimal places.[4]
Show the answer and mark scheme
Answer: \(x = 0.921, \ 2.744\)
M1 for \(\sin\left(2x + \frac{\pi}{3}\right) = \frac{1}{4}\) and a principal value \(2x + \frac{\pi}{3} = 0.2526\ldots\)
A1 for one correct value of \(x\)
dM1 for a correct method to find a further value of \(2x + \frac{\pi}{3}\) in the interval \(\frac{\pi}{3} \le 2x + \frac{\pi}{3} \lt \frac{7\pi}{3}\) and solving for \(x\)
A1 for all of \(0.921, \ 2.744\) and no others in the interval
Worked solution: \(\sin\left(2x + \frac{\pi}{3}\right) = \frac{1}{4}\), with \(\frac{\pi}{3} \le 2x + \frac{\pi}{3} \lt \frac{7\pi}{3}\). \(2x + \frac{\pi}{3} = 2.8889\ldots, \ 6.5358\ldots\) so \(x = 0.921, \ 2.744\).
Question 3Hard10 marks
The function \(\mathrm{f}\) is defined by \[\mathrm{f}(x) = 3\arcsin(2x - 1)\] for its largest possible domain.
(a) Find the domain and the range of \(\mathrm{f}\).[3]
(b) Find \(\mathrm{f}^{-1}(x)\), stating its domain.[4]
(c) Solve the equation \(\mathrm{f}(x) = -\frac{\pi}{2}\).[3]
Show the answer and mark scheme
(a)Answer: Domain \(0 \le x \le 1\); range \(-\frac{3\pi}{2} \le \mathrm{f}(x) \le \frac{3\pi}{2}\)
M1 for \(-1 \le 2x - 1 \le 1\)
A1 for domain \(0 \le x \le 1\)
B1 for range \(-\frac{3\pi}{2} \le \mathrm{f}(x) \le \frac{3\pi}{2}\)
Worked solution: \(-1 \le 2x - 1 \le 1 \Rightarrow 0 \le x \le 1\). The range of \(\arcsin\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so \(-\frac{3\pi}{2} \le \mathrm{f}(x) \le \frac{3\pi}{2}\).
M1 for making \(\arcsin(2x - 1)\) the subject of \(y = 3\arcsin(2x - 1)\)
M1 for taking \(\sin\) of both sides and rearranging for \(x\)
A1 for \(\mathrm{f}^{-1}(x) = \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) + 1\right)\) oe
B1ft for domain \(-\frac{3\pi}{2} \le x \le \frac{3\pi}{2}\) (the range of \(\mathrm{f}\))
Worked solution: \(y = 3\arcsin(2x - 1) \Rightarrow \arcsin(2x - 1) = \frac{y}{3}\) \(\Rightarrow 2x - 1 = \sin\left(\frac{y}{3}\right)\), so \(\mathrm{f}^{-1}(x) = \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) + 1\right)\). The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\): \(-\frac{3\pi}{2} \le x \le \frac{3\pi}{2}\).
(c)Answer: \(x = \frac{1}{4}\)
M1 for \(\arcsin(2x - 1) = -\frac{\pi}{6}\)
M1 for \(2x - 1 = \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\)