1. **Problem statement:** Given $\sin \alpha = \frac{3}{5}$ with $\alpha$ in the first quadrant, find: - 5.3.1 $\tan \alpha$ - 5.3.2 $\sin \beta$ - 5.3.3 Coordinates of point $P$ Then, prove the identity: $$\frac{\sin \theta - \tan \theta \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta} = \tan \theta$$ --- 2. **Step 5.3.1: Find $\tan \alpha$ given $\sin \alpha = \frac{3}{5}$** - Since $\alpha$ is in the first quadrant, all trigonometric values are positive. - Using Pythagoras theorem in the right triangle: $$\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$$ - Find adjacent side: $$\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$ - Then, $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$$ --- 3. **Step 5.3.2: Find $\sin \beta$** - Given $OP = 2 RO$ and $\angle TOP = \beta$. - Point $P$ lies on the line $RO$ extended, so the vector length doubles and $P$ lies in the opposite direction of $R$ relative to $O$ (third quadrant). - Since $\beta$ is angle between $OP$ and positive x-axis and $\beta$ is in the third quadrant, - Coordinates of $R$ can be found from $\alpha$: $$R = (x, y) = (\cos \alpha, \sin \alpha) = \left(\frac{4}{5}, \frac{3}{5}\right)$$ - Vector $OP$ is twice $RO$ but in opposite direction, so: $$P = -2R = \left(-2\times \frac{4}{5}, -2 \times \frac{3}{5}\right) = \left(-\frac{8}{5}, -\frac{6}{5}\right)$$ - Then, $$\sin \beta = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{-\frac{6}{5}}{|P|}$$ - Length $|P| = 2 |R| = 2 \times 1 = 2$ (since $R$ lies on unit circle basis) - Hence: $$\sin \beta = \frac{-\frac{6}{5}}{2} = -\frac{6}{10} = -\frac{3}{5}$$ --- 4. **Step 5.3.3: Coordinates of $P$** - As from above: $$P = \left(-\frac{8}{5}, -\frac{6}{5}\right)$$ --- 5. **Proof of identity:** Prove: $$\frac{\sin \theta - \tan \theta \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta} = \tan \theta$$ - Start with left-hand side (LHS): $$\text{LHS} = \frac{\sin \theta - \tan \theta \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta}$$ - Replace $\tan \theta = \frac{\sin \theta}{\cos \theta}$: $$= \frac{\sin \theta - \frac{\sin \theta}{\cos \theta} \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta} = \frac{\sin \theta - \sin \theta \cos \theta}{\cos \theta - 1 + \sin^2 \theta}$$ - Factor numerator: $$= \frac{\sin \theta(1 - \cos \theta)}{\cos \theta - 1 + \sin^2 \theta}$$ - Use Pythagorean identity $\sin^2 \theta = 1 - \cos^2 \theta$ to simplify the denominator: $$\cos \theta - 1 + (1 - \cos^2 \theta) = \cos \theta - 1 + 1 - \cos^2 \theta = \cos \theta - \cos^2 \theta$$ - Factor denominator: $$= \cos \theta (1 - \cos \theta)$$ - So LHS becomes: $$\frac{\sin \theta (1 - \cos \theta)}{\cos \theta (1 - \cos \theta)}$$ - Cancel $(1 - \cos \theta)$ (nonzero for $\theta \neq 0$): $$= \frac{\sin \theta}{\cos \theta} = \tan \theta$$ - Therefore, identity proven: $$\boxed{\frac{\sin \theta - \tan \theta \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta} = \tan \theta}$$ --- **Final answers:** - 5.3.1 $\tan \alpha = \frac{3}{4}$ - 5.3.2 $\sin \beta = -\frac{3}{5}$ - 5.3.3 $P = \left(-\frac{8}{5}, -\frac{6}{5}\right)$ - 5.4 Identity is proven as shown.