Trigonometric graphs
Geometry and measures
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Complete $\sin 0°$, $\sin 90°$, $\sin 180°$ and $\sin 270°$.
The graph of y = sin x from 0° to 360° 2 marks
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Write the values of $\cos 0°$, $\cos 60°$ and $\cos 180°$.
2 marks
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Solve $\sin x = \dfrac{1}{2}$ for $0° \le x \le 360°$.
3 marks
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Solve $\cos x = \dfrac{1}{2}$ for $0° \le x \le 360°$.
3 marks
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From the graph of $y = \sin x$, write the maximum value and where it occurs, the minimum value and where it occurs, and the other angle between $0°$ and $360°$ where $\sin x = \sin 30°$.
4 marks
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Solve $\cos x = -\dfrac{1}{2}$ for $0° \le x \le 360°$.
4 marks
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The graph of $y = \sin x$ is changed to $y = 2\sin x$. Write the new amplitude and solve $2\sin x = 1$ for $0° \le x \le 360°$. Describe the transformation from $y = \sin x$ to $y = \sin(x - 90°)$.
5 marks
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Solve $\tan x = 1$ for $0° \le x \le 360°$. Then solve $2\sin x + \sqrt{3} = 0$ in the same interval.
6 marks
Mark scheme
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Answer. $0, 1, 0, -1$
- The sine graph starts at $0$, reaches $1$ at $90°$ and returns to $0$ at $180°$.
- At $270°$ the value is $-1$.
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Answer. $1, \dfrac{1}{2}, -1$
- $\cos 0° = 1$ and $\cos 60° = \dfrac{1}{2}$.
- $\cos 180° = -1$
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Answer. $x = 30°$ or $x = 150°$
- The first solution is $30°$.
- The second is $180 - 30 = 150°$.
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Answer. $x = 60°$ or $x = 300°$
- The first solution is $60°$.
- The second is $360 - 60 = 300°$.
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Answer. Maximum $1$ at $90°$, minimum $-1$ at $270°$, and $150°$
- The maximum is $1$ at $90°$.
- The minimum is $-1$ at $270°$.
- $\sin 150° = \sin 30°$.
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Answer. $x = 120°$ or $x = 240°$
- The related acute angle is $60°$.
- In the second and third quadrants, $x = 120°$ or $x = 240°$.
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Answer. Amplitude $2$. Solutions $30°$ and $150°$. A translation of $90°$ to the right.
- The stretch parallel to the $y$-axis has scale factor $2$.
- $\sin x = \dfrac{1}{2}$, so $x = 30°$ or $x = 150°$.
- $y = \sin(x - 90°)$ is a translation by $90°$ to the right.
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Answer. $\tan x = 1$ at $45°$ and $225°$. $2\sin x + \sqrt{3} = 0$ at $240°$ and $300°$.
- $\tan x = 1$ at $45°$, and the tangent repeats after $180°$, so also at $225°$.
- $2\sin x = -\sqrt{3}$, so $\sin x = -\dfrac{\sqrt{3}}{2}$.
- The solutions are $240°$ and $300°$.