Find \(\displaystyle \int \sin 3x \, \cos x \, dx\). (2 marks)
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Calculus, EXT1 C2 2020 HSC 12d
Find `int_0^(pi/2) cos 5x\ sin 3x\ dx`. (3 marks)
Calculus, EXT1 C2 2019 HSC 14c
The diagram shows the two curves `y = sin x` and `y = sin(x - alpha) + k`, where `0 < alpha < pi` and `k > 0`. The two curves have a common tangent at `x_0` where `0 < x_0 < pi/2`.
- Explain why `cos x_0 = cos (x_0 - alpha)`. (1 mark)
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- Show that `sin x_0 = -sin(x_0 - alpha)`. (2 marks)
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- Hence, or otherwise, find `k` in terms of `alpha`. (2 marks)
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Show Worked Solution
| i. | `y_1` | `= sin x` |
| `(dy_1)/(dx)` | `= cos x` | |
| `y_2` | `= sin(x – alpha) + k` | |
| `(dy_2)/(dx)` | `= cos (x – alpha)` |
`text(At)\ \ x = x_0,\ \ text(tangent is common)`♦ Mean mark part (i) 47%.
`:. cos x_0 = cos(x_0 – alpha)`
ii. `x_0\ \ text{is in 1st quadrant (given)}`
`text{Using part (i):}`
`cos\ x_0 = cos(x_0 – alpha) >0`♦♦♦ Mean mark part (ii) 19%.
`=> x_0 – alpha\ \ \ text(is in 4th quadrant)\ \ (0 < alpha < pi)`
`text(S)text(ince sin is positive in 1st quadrant and)`
`text(negative in 4th quadrant)`
`=> sin x_0 = -sin(x_0 – alpha)`
| iii. |  |
`text(When)\ \ x = x_0,`
| `y_1` | `=sin x_0` | |
| `y_2` | `=sin(x_0 – alpha) + k` | |
| `sin x_0` | `=sin (x_0 – alpha) + k` | |
| `= -sin x_0 + k` | ||
| `k` | `== 2\ sin x_0` |
♦♦ Mean mark part (iii) 21%.
| `text(S)text(ince)\ \ cos x_0` | `= cos(x_0 – alpha)` |
| `x_0` | `= -(x_0 – alpha)` |
| `2x_0` | `= alpha` |
| `x_0` | `= alpha/2` |
`:. k = 2 sin\ alpha/2`
Calculus, EXT1 C2 2005 HSC 3b
- By expanding the left-hand side, show that
- `qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)` (1 mark)
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- Hence find `int sin(5x) cos (4x)\ dx.` (2 marks)
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