P5.8Proving trigonometric identities

• M1 for \(1 - \sin^2 {x} = \cos^2 {x}\) and \(1 + \tan^2 {x} = \sec^2 {x}\)
• A1* for \(\cos^2 {x}\sec^2 {x} = 1\)

Worked solution: \((1 - \sin^2 {x})(1 + \tan^2 {x}) = \cos^2 {x} \times \sec^2 {x} = \cos^2 {x} \times \frac{1}{\cos^2 {x}} = 1\)

• M1 for expanding both: \(\sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B\)
• A1* for \(2\sin A\cos B\)

Worked solution: \(\sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B = 2\sin A\cos B\)

• M1 for writing as \(\frac{1 + \cos 2\theta}{\sin 2\theta}\)
• M1 for using double angle formulae: \(\frac{2\cos^2 \theta}{2\sin \theta\cos \theta}\)
• A1* for \(\frac{\cos \theta}{\sin \theta} = \cot \theta\)

Worked solution: \(\operatorname{cosec} 2\theta + \cot 2\theta = \frac{1 + \cos 2\theta}{\sin 2\theta} = \frac{2\cos^2 \theta}{2\sin \theta\cos \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

• M1 for using part (a) to obtain \(\cot \theta = -\frac{1}{3} \Rightarrow \tan \theta = -3\)
• A1 for one correct solution
• dM1 for a correct method to find a second solution in the interval
• A1 for \(-71.6^\circ, \ 108.4^\circ\) and no others

Worked solution: By part (a), \(\cot \theta = -\frac{1}{3} \Rightarrow \tan \theta = -3\).
The solutions in the interval are \(\theta = -71.6^\circ, \ 108.4^\circ\).

• M1 for \(\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta\)
• M1 for \(2\sin^2\theta\cos^2\theta = \frac{1}{2}\sin^2 2\theta\)
• M1 for \(\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}\)
• A1* for a complete proof

Worked solution: \(\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}\sin^2 2\theta\)
\(= 1 - \frac{1}{2} \times \frac{1 - \cos 4\theta}{2} = \frac{4 - 1 + \cos 4\theta}{4} = \frac{3 + \cos 4\theta}{4}\)

• B1 for either end value
• B1 for \(\frac{1}{2} \le \mathrm{f}(\theta) \le 1\)

Worked solution: Since \(-1 \le \cos 4\theta \le 1\), \(\frac{2}{4} \le \mathrm{f}(\theta) \le \frac{4}{4}\), i.e. \(\frac{1}{2} \le \mathrm{f}(\theta) \le 1\).

• M1 for \(\frac{3 + \cos 4\theta}{4} = \frac{5}{8} \Rightarrow \cos 4\theta = -\frac{1}{2}\)
• A1 for \(4\theta = 120^\circ\) (or \(\theta = 30^\circ\))
• M1 for finding further values of \(4\theta\) in \(0 \le 4\theta \lt 720^\circ\)
• A1 for all four values and no others

Worked solution: \(\cos 4\theta = \frac{5}{2} - 3 = -\frac{1}{2}\), with \(0 \le 4\theta \lt 720^\circ\): \(4\theta = 120^\circ, 240^\circ, 480^\circ, 600^\circ\), so \(\theta = 30^\circ, 60^\circ, 120^\circ, 150^\circ\).