June 2017 Paper 2 Q19
19 Here is a sketch of \(\quad y = \sin x \quad\) for \(\quad 0^\circ \leqslant x \leqslant 360^\circ\)
Not drawn accurately
\(\alpha\) is an acute angle measured in degrees.
\(\sin\alpha = k \qquad\) where \(k\) is a constant.
Write the answers to each of the following in terms of \(k\), without involving trigonometric functions.
(a) \(\sin(180^\circ - \alpha)\) [1 mark]
(b) \(\sin(360^\circ - \alpha)\) [1 mark]
(c) \(\cos\alpha\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(k\) | B1 |
Additional guidance
| \(k = 0\) or \(k = 1\) etc | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(-k\) | B1 |
Additional guidance
| \(-k = 0\) or \(-k = 1\) etc | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(k^2 + \cos^2\alpha = 1\) or \(1 - k^2\) | M1 | oe eg \((1 + k)(1 - k)\) |
| \(\sqrt{1 - k^2}\) or \(\sqrt{(1 + k)(1 - k)}\) | A1 |
Additional guidance
| Answer \(-\sqrt{1 - k^2}\) or \(\pm\sqrt{1 - k^2}\) | M1A0 |
| Correct answer followed by incorrect further work | M1A0 |
| Answer \(1 - k^2\) | M1A0 |
| Allow \(\cos^2 x\) or \(\cos^2\theta\) etc or \(\cos^2\) or c\(^2\) or \((\cos\alpha)^2\) for \(\cos^2\alpha\) | |
| Condone \(\cos\alpha^2\) for \(\cos^2\alpha\) | |
| \(\cos(\sin^{-1} k)\) | M0A0 |