Trigonometric graphs and equations
Trigonometry
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Solve $\sin x = \dfrac{1}{2}$ for $0^\circ \le x < 360^\circ$.
3 marks
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State the period of $y = \cos 2x$, and the maximum value of $y = 3\sin x - 1$.
3 marks
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Solve $2\cos x + 1 = 0$ for $0 \le x \le 2\pi$.
4 marks
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Describe the two transformations that map $y = \sin x$ onto $y = 2\sin\left(x - \dfrac{\pi}{6}\right)$.
4 marks
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Solve $\sin 2x = \dfrac{\sqrt{3}}{2}$ for $0^\circ \le x \le 360^\circ$.
5 marks
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Prove that $\dfrac{1 - \cos^2 x}{\sin x \cos x} = \tan x$, for values of $x$ where the expression is defined.
5 marks
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Solve $2\sin^2 x - \sin x - 1 = 0$ for $0^\circ \le x < 360^\circ$.
6 marks
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Solve $\cos 2x = \sin x$ for $0 \le x \le 2\pi$.
7 marks
Mark scheme
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Answer. $x = 30^\circ$ or $x = 150^\circ$
- The principal value is $30^\circ$.
- The second solution in the interval is $180^\circ - 30^\circ$.
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Answer. The period is $180^\circ$, and the maximum value is $2$.
- The period of $\cos 2x$ is $\dfrac{360^\circ}{2}$.
- The maximum of $3\sin x$ is $3$, so the maximum of $3\sin x - 1$ is $2$.
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Answer. $x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$
- $\cos x = -\dfrac{1}{2}$.
- The solutions in the interval are $\dfrac{2\pi}{3}$ and $\dfrac{4\pi}{3}$.
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Answer. A translation by $\begin{pmatrix} \dfrac{\pi}{6} \\ 0 \end{pmatrix}$, followed by a stretch of scale factor $2$ parallel to the $y$-axis.
- Replacing $x$ by $x - \dfrac{\pi}{6}$ translates the graph by $\dfrac{\pi}{6}$ in the positive $x$-direction.
- The factor $2$ then stretches the graph parallel to the $y$-axis.
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Answer. $x = 30^\circ$, $60^\circ$, $210^\circ$ or $240^\circ$
- $2x$ lies in $[0^\circ, 720^\circ]$.
- $2x = 60^\circ$, $120^\circ$, $420^\circ$ or $480^\circ$.
- Divide by $2$.
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Answer. The identity is true.
- The numerator is $\sin^2 x$.
- $\dfrac{\sin^2 x}{\sin x \cos x} = \dfrac{\sin x}{\cos x} = \tan x$.
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Answer. $x = 90^\circ$, $210^\circ$ or $330^\circ$
- $(2\sin x + 1)(\sin x - 1) = 0$.
- $\sin x = 1$ gives $x = 90^\circ$.
- $\sin x = -\dfrac{1}{2}$ gives $x = 210^\circ$ and $x = 330^\circ$.
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Answer. $x = \dfrac{\pi}{6}$, $\dfrac{5\pi}{6}$ or $\dfrac{3\pi}{2}$
- Use $\cos 2x = 1 - 2\sin^2 x$.
- $2\sin^2 x + \sin x - 1 = 0$, so $(2\sin x - 1)(\sin x + 1) = 0$.
- $\sin x = \dfrac{1}{2}$ gives $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$.
- $\sin x = -1$ gives $\dfrac{3\pi}{2}$.