June 2016 Paper 2 Q18
18 Angle \(y\) is acute.
\[\tan y = \frac{p + 1}{p - 1} \qquad \text{where } p \text{ is a constant greater than 1}\](a) Which of the statements below is correct?
Circle your answer. [1 mark]
- \(y = 45^\circ\)
- \(y \lt 45^\circ\)
- \(y \gt 45^\circ\)
- \(y\) could be any acute angle
(b) Work out the expression for \(\sin y\)
Give your answer in the form \(\ \dfrac{ap + b}{\sqrt{cp^2 + d}}\ \) where \(a\), \(b\), \(c\) and \(d\) are integers.
You may use a diagram to help you. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(y \gt 45^\circ\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \((p + 1)^2 + (p - 1)^2\) | M1 | oe May be within a square root |
| \(p^2 + p + p + 1 + p^2 - p - p + 1\) or \(p^2 + 1 + p^2 + 1\) | M1 | oe May be within a square root Implies M1 M1 |
| \(2p^2 + 2\) or \(2(p^2 + 1)\) | A1 | May be within a square root Must be simplified May be implied by final mark |
| \(\dfrac{p + 1}{\sqrt{2p^2 + 2}}\) | A1 | Allow \(a = b = 1 \quad c = d = 2\) SC2 \(\dfrac{p + 1}{\sqrt{(p + 1)^2 + (p - 1)^2}}\) |
Additional guidance
| \(\dfrac{p + 1}{\sqrt{2p^2 + 2}}\) and further incorrect work | M2 A1 A0 |
| Allow \(1p\) for \(p\) | |
| Use of \(\dfrac{\sin y}{\cos y} = \dfrac{p + 1}{p - 1}\) can be marked by the scheme eg \((p - 1)^2\sin^2 y = (p + 1)^2\cos^2 y\) \((p - 1)^2\sin^2 y = (p + 1)^2(1 - \sin^2 y)\) \(((p - 1)^2 + (p + 1)^2)\sin^2 y = (p + 1)^2\) First M1 gained here (M1 A1 A1 may subsequently be gained) |