Chapter 6: Trigonometric Identities LO3 AS11.3.5 8. \( f(\theta)=\sqrt{2} \cos \theta+\sqrt{2} \sin \theta \) a) Write \( f(\theta) \) in the form: \( f(\theta)=A \sin (\varphi+\theta) \). b) (i) Hence find the maximum value of \( f(\theta) \) and the values of \( \theta \) at which it occurs. (ii) Solve \( f(\theta)=1 \). 9. Solve: \( \cos \theta+\cos \left(\frac{\theta}{2}\right)+1=0, \theta \in\left(-90^{\circ} ; 90^{\circ}\right) \). 111). Solve: \( 5 \cos ^{2}\left(\frac{\theta}{2}\right)+\sin \theta=3 \sin ^{2}\left(\frac{\theta}{2}\right)+2, \theta \in\left(-360^{\circ} ; 360^{\circ}\right) \). 11. Solve the following equations: a) \( \sqrt{1+\sin \theta}=\cos \theta, \theta \in\left(0^{\circ} ; 360^{\circ}\right) \) b) \( \frac{\sqrt{1+\cos \theta}}{2}=\cos \theta, \theta \in\left(-360^{\circ} ; 0^{\circ}\right) \) (c) \( \sin \theta=\sqrt{2-\sin \theta}, \theta \in\left(-90^{\circ} ; 180^{\circ}\right] \)

Let's solve the problems step by step. ### Problem 8 Given the function: \[ f(\theta) = \sqrt{2} \cos \theta + \sqrt{2} \sin \theta \] #### Part a We want to express \( f(\theta) \) in the form: \[ f(\theta) = A \sin(\varphi + \theta) \] To do this, we can use the sine addition formula: \[ \sin(\varphi + \theta) = \sin \varphi \cos \theta + \cos \varphi \sin \theta \] We can match coefficients: - \( A \sin \varphi = \sqrt{2} \) - \( A \cos \varphi = \sqrt{2} \) To find \( A \), we can use the Pythagorean identity: \[ A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \] Now, we can find \( \varphi \): \[ \sin \varphi = \frac{\sqrt{2}}{2}, \quad \cos \varphi = \frac{\sqrt{2}}{2} \] This gives us \( \varphi = 45^\circ \) or \( \varphi = \frac{\pi}{4} \). Thus, we can write: \[ f(\theta) = 2 \sin\left(\frac{\pi}{4} + \theta\right) \] #### Part b (i) The maximum value of \( f(\theta) \) occurs when \( \sin\left(\frac{\pi}{4} + \theta\right) = 1 \): \[ 2 \cdot 1 = 2 \] To find the values of \( \theta \): \[ \frac{\pi}{4} + \theta = \frac{\pi}{2} + 2k\pi \quad \Rightarrow \quad \theta = \frac{\pi}{4} + 2k\pi \] \[ \frac{\pi}{4} + \theta = \frac{3\pi}{2} + 2k\pi \quad \Rightarrow \quad \theta = \frac{5\pi}{4} + 2k\pi \] (ii) To solve \( f(\theta) = 1 \): \[ 2 \sin\left(\frac{\pi}{4} + \theta\right) = 1 \quad \Rightarrow \quad \sin\left(\frac{\pi}{4} + \theta\right) = \frac{1}{2} \] The solutions are: \[ \frac{\pi}{4} + \theta = \frac{\pi}{6} + 2k\pi \quad \Rightarrow \quad \theta = \frac{\pi}{6} - \frac{\pi}{4} + 2k\pi = -\frac{\pi}{12} + 2k\pi \] \[ \frac{\pi}{4} + \theta = \frac{5\pi}{6} + 2k\pi \quad \Rightarrow \quad \theta = \frac{5\pi}{6} - \frac{\pi}{4} + 2k\pi = \frac{7\pi}{12} + 2k\pi \] ### Problem 9 Solve: \[ \cos \theta + \cos\left(\frac{\theta}{2}\right) + 1 = 0 \] Rearranging gives: \[ \cos \theta + 1 = -\cos\left(\frac{\theta}{2}\right) \] Using the identity \( \cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \): \[ 2\cos^2\left(\frac{\theta}{2}\right) - 1 + 1 = -\cos\left(\frac{\theta}{2}\right) \] This simplifies to: \[ 2\cos^2\left(\frac{\theta}{2}\right) = -\cos\left(\frac{\theta}{2}\right) \] Factoring gives: \[ \cos\left(\frac{\theta}{2}\right)(2\cos\left(\frac{\theta}{2}\right) + 1) = 0 \] Thus, we have: 1. \( \cos\left(\frac{\theta}{2}\right) = 0 \) leads to \( \frac{\theta}{2} = 90^\circ + k \cdot 180^\circ \) or \( \theta = 180^\circ + k \cdot 360^\circ \) 2. \( 2\cos\left(\frac{\theta}{2}\right) + 1 = 0 \) leads to \( \cos\left(\frac{\theta}{2}\right) = -\frac{1}{2} \) or \( \frac{\theta}{2} = 120^\circ + k \cdot 360^\circ \) or \( \frac{\theta}{2} = 240^\circ + k \cdot 360^\circ \) ### Problem 11 #### Part a Solve: \[ \sqrt{1 + \sin \theta} = \cos \theta \] Squaring both sides: \[ 1 + \sin \theta = \cos^2 \theta \] Using \( \cos^2 \theta = 1 - \sin^2 \theta \): \[ 1 + \sin \theta = 1 - \sin^2 \theta \] Rearranging gives: \[ \sin^2 \theta + \sin \theta = 0 \] Factoring: \[ \sin \theta(\sin \theta + 1) = 0 \] Thus, \( \sin \theta = 0 \) or \( \sin \theta = -1 \): - \( \theta = 0^\circ, 180^\circ \) - \( \theta = 270^\circ \) #### Part b Solve: \[ \frac{\sqrt{1 + \cos \theta}}{2} = \cos \theta \] Squaring both sides: \[ \frac{1 + \cos \theta}{4} = \cos^2 \theta \] Rearranging gives: \[ 1 + \cos \theta = 4\cos^2 \theta \] This leads to: \[ 4\cos^2 \theta - \cos \theta - 1 = 0 \] Using the quadratic formula: \[ \cos \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{1 \pm