18(a) Show that \(\cos 3\theta \equiv 4 \cos^3 \theta - 3 \cos \theta\). 1. Start with the triple-angle formula for cosine: $$\cos 3\theta = \cos(2\theta + \theta)$$ 2. Use cosine addition formula: $$\cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta$$ 3. Substitute double-angle formulas: $$\cos 2\theta = 2\cos^2 \theta - 1, \quad \sin 2\theta = 2\sin \theta \cos \theta$$ 4. Write: $$\cos 3\theta = (2\cos^2 \theta - 1)\cos \theta - 2\sin \theta \cos \theta \sin \theta$$ 5. Simplify: $$= 2\cos^3 \theta - \cos \theta - 2\cos \theta \sin^2 \theta$$ 6. Use identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to get: $$= 2\cos^3 \theta - \cos \theta - 2\cos \theta (1 - \cos^2 \theta)$$ 7. Simplify inside parentheses: $$= 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta$$ 8. Combine like terms: $$= 4\cos^3 \theta - 3\cos \theta$$ 18(b) Solve for \(-\pi \leq \theta \leq \pi\): $$1 - \cos 3x = \sin^2 x$$ 1. Recall identity \(\sin^2 x = 1 - \cos^2 x\). 2. Rewrite equation: $$1 - \cos 3x = 1 - \cos^2 x$$ 3. Simplify: $$-\cos 3x = -\cos^2 x$$ 4. Multiply both sides by -1: $$\cos 3x = \cos^2 x$$ 5. Using previous result, write \(\cos 3x = 4\cos^3 x - 3\cos x\): $$4\cos^3 x - 3\cos x = \cos^2 x$$ 6. Rearrange: $$4\cos^3 x - \cos^2 x - 3\cos x = 0$$ 7. Factor out \(\cos x\): $$\cos x (4\cos^2 x - \cos x - 3) = 0$$ 8. Solve for each factor: - \(\cos x = 0\) - Quadratic in \(\cos x\): \(4u^2 - u - 3 = 0\) where \(u=\cos x\) 9. Solve quadratic: $$u = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}$$ 10. Roots: $$u = 1$$ $$u = -\frac{3}{4}$$ 11. Find corresponding \(x\) values in \([-\pi, \pi]\): - \(\cos x = 0\) \(\Rightarrow x = -\frac{\pi}{2}, \frac{\pi}{2}\) - \(\cos x = 1\) \(\Rightarrow x = 0\) - \(\cos x = -\frac{3}{4}\) \(\Rightarrow x = \pm \arccos\left(-\frac{3}{4}\right)\) 19(a) Transformations mapping \(y = \sin x\) to \(y = \sqrt{3} \sin x - \cos x + 4\): 1. Recognize expression combines sine and cosine: $$y = \sqrt{3}\sin x - \cos x + 4 = R \sin(x + \alpha) + 4$$ where $$R = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2$$ 2. Find phase shift \(\alpha\): $$\tan \alpha = \frac{-1}{\sqrt{3}} = -\frac{1}{\sqrt{3}} \Rightarrow \alpha = -\frac{\pi}{6}$$ 3. So, $$y = 2 \sin\left(x - \frac{\pi}{6}\right) + 4$$ 4. Transformations: - Vertical stretch by 2 - Phase shift (horizontal translation) right by \(\frac{\pi}{6}\) - Vertical translation up by 4 19(b) Minimum value of: $$\frac{1}{\sqrt{3} \sin x - \cos x + 4} = \frac{1}{2 \sin(x - \frac{\pi}{6}) + 4}$$ 1. Minimum denominator occurs at minimum of: $$2\sin (x - \frac{\pi}{6}) + 4$$ 2. Minimum of \(\sin\) is \(-1\): $$\min = 2(-1) + 4 = 2$$ 3. Therefore minimum value of original expression: $$\frac{1}{2}$$ 20(a) Show $$\frac{1 - \cos 2x}{1 + \cos 2x} \equiv \tan^2 x$$ 1. Recall identities: $$\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$$ 2. Substitute numerator: $$1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x$$ 3. Substitute denominator: $$1 + \cos 2x = 1 + (1 - 2\sin^2 x) = 2 - 2\sin^2 x = 2\cos^2 x$$ 4. Therefore: $$\frac{1 - \cos 2x}{1 + \cos 2x} = \frac{2\sin^2 x}{2\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$ 20(b) Solve $$\frac{1 - \cos 2x}{1 + \cos 2x} = 3$$ 1. From (a), replace left side by \(\tan^2 x\): $$\tan^2 x = 3$$ 2. So, $$\tan x = \pm \sqrt{3}$$ 3. For \(-\pi \leq x \leq \pi\), values where tangent equals \(\sqrt{3}\) are: $$x = \pm \frac{\pi}{3}$$ 4. Values where tangent equals \(-\sqrt{3}\) are: $$x = \pm \frac{2\pi}{3}$$ 21(a) Prove identity: $$\frac{1 - \cos 2x}{\sin 2x} \equiv \tan x$$ 1. Use identities: $$1 - \cos 2x = 2\sin^2 x\quad \text{and} \quad \sin 2x = 2\sin x \cos x$$ 2. Substitute: $$\frac{1 - \cos 2x}{\sin 2x} = \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$$ 21(b) Solve for \(0 \leq \theta < 2\pi\): $$1 - \cos 2\theta = \sin 2\theta$$ 1. Rearrange: $$1 - \cos 2\theta - \sin 2\theta = 0$$ 2. Use sum-to-product formulas or rewrite: $$1 = \cos 2\theta + \sin 2\theta$$ 3. Express RHS as single sinusoidal: $$\cos 2\theta + \sin 2\theta = \sqrt{2} \sin(2\theta + \frac{\pi}{4})$$ 4. Equation becomes: $$1 = \sqrt{2} \sin(2\theta + \frac{\pi}{4})$$ 5. So: $$\sin(2\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$$ 6. Solutions for \(0 \leq \theta < 2\pi\) come from: $$2\theta + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad \pi - \frac{\pi}{4} + 2k\pi$$ 7. Simplify: - $$2\theta = 2k\pi$$ gives $$\theta = k\pi$$ - $$2\theta = \frac{3\pi}{4} + 2k\pi$$ gives $$\theta = \frac{3\pi}{8} + k\pi$$ 8. Valid \(\theta\) in interval: $$\theta = 0, \pi, \frac{3\pi}{8}, \frac{11\pi}{8}$$ Slug: "trig identities and solves" Subject: "trigonometry" Desmos: {"latex":"y=\sqrt{3} \sin x - \cos x + 4","features":{"intercepts":true,"extrema":true}} q_count:8