Let's go through each problem step by step, showing the formulas and process used. --- ### 19. Convert the following Cartesian equation into a polar Equation: \( x^2 - y^2 = 3y \) **Formulas:** - \( x = r \cos \theta \) - \( y = r \sin \theta \) **Steps:** \[ x^2 - y^2 = 3y \] Substitute \( x \) and \( y \) in terms of \( r \) and \( \theta \): \[ (r\cos\theta)^2 - (r\sin\theta)^2 = 3(r\sin\theta) \] \[ r^2 \cos^2\theta - r^2 \sin^2\theta = 3r\sin\theta \] \[ r^2 (\cos^2\theta - \sin^2\theta) = 3r\sin\theta \] If \( r \neq 0 \), divide both sides by \( r \): \[ r (\cos^2\theta - \sin^2\theta) = 3\sin\theta \] --- ### 20. Convert the polar equation into a Cartesian equation: \( \theta = \frac{\pi}{4} \) **Formula:** \( \tan \theta = \frac{y}{x} \) **Steps:** \[ \theta = \frac{\pi}{4} \] \[ \tan\theta = 1 \implies \frac{y}{x} = 1 \implies y = x \] --- ### 21. Give two more polar coordinates that describe the point \( (-3, \pi) \) **Formula:** \( (r, \theta) \) and \( (-r, \theta+\pi) \) describe the same point. **Steps:** - The given point is \( (r, \theta) = (-3, \pi) \) - Two more representations: 1. \( (3, 0) \) (since \( -3 \) at \( \pi \) is the same as \( 3 \) at \( 0 \)) 2. \( (3, 2\pi) \) (since angle can differ by any multiple of \( 2\pi \)) --- ### 22. Find the exact value of: #### (a) \( \sin^{-1}(-\frac{1}{2}) \) We know \( \sin(\theta) = -\frac{1}{2} \) at \( \theta = -\frac{\pi}{6} \) (principal value). **Answer:** \( -\frac{\pi}{6} \) #### (b) \( \cos^{-1}(-\frac{1}{2}) \) We know \( \cos(\theta) = -\frac{1}{2} \) at \( \theta = \frac{2\pi}{3} \) (principal value). **Answer:** \( \frac{2\pi}{3} \) #### (c) \( \cos^{-1}(-\frac{\sqrt{2}}{2}) \) We know \( \cos(\theta) = -\frac{\sqrt{2}}{2} \) at \( \theta = \frac{3\pi}{4} \). **Answer:** \( \frac{3\pi}{4} \) #### (d) \( \tan^{-1}(-1) \) We know \( \tan(\theta) = -1 \) at \( \theta = -\frac{\pi}{4} \). **Answer:** \( -\frac{\pi}{4} \) --- ### 23. Find the exact value of \( \cos(\tan^{-1}(4) - \sin^{-1}(\frac{5}{9})) \) Let \( A = \tan^{-1}(4) \), \( B = \sin^{-1}(\frac{5}{9}) \). **Formula:** \( \cos(A-B) = \cos A \cos B + \sin A \sin B \) - For \( A \): \( \tan A = 4 \Rightarrow \) Opposite = 4, Adjacent = 1, Hypotenuse = \( \sqrt{17} \) - \( \sin A = \frac{4}{\sqrt{17}} \), \( \cos A = \frac{1}{\sqrt{17}} \) - For \( B \): \( \sin B = \frac{5}{9} \Rightarrow \) Opposite = 5, Hypotenuse = 9, Adjacent = \( \sqrt{81-25} = \sqrt{56} \) - \( \cos B = \frac{\sqrt{56}}{9} \), \( \sin B = \frac{5}{9} \) Now, \[ \cos(A-B) = \cos A \cos B + \sin A \sin B \] \[ = \left(\frac{1}{\sqrt{17}}\right)\left(\frac{\sqrt{56}}{9}\right) + \left(\frac{4}{\sqrt{17}}\right)\left(\frac{5}{9}\right) \] \[ = \frac{\sqrt{56}}{9\sqrt{17}} + \frac{20}{9\sqrt{17}} \] \[ = \frac{\sqrt{56} + 20}{9\sqrt{17}} \] --- ### 24. If \( \csc 2\theta = -\frac{14}{9} \) and \( -\pi \leq 2\theta \leq 0 \), find exact values for the six trigonometric functions of \( \theta \). - \( \csc 2\theta = -\frac{14}{9} \implies \sin 2\theta = -\frac{9}{14} \) - Use the double angle formulas: Let \( 2\theta = \alpha \), \( \sin \alpha = -\frac{9}{14} \) - \( \cos^2 \alpha = 1 - \sin^2\alpha = 1 - \frac{81}{196} = \frac{115}{196} \) - \( \cos\alpha = -\frac{\sqrt{115}}{14} \) (pick negative since \( 2\theta \in [-\pi, 0] \), so \( 2\theta \) is in third or fourth quadrant, where cosine is negative) Let \( \sin 2\theta = 2\sin\theta\cos\theta = -\frac{9}{14} \) Let \( \sin\theta = x \), \( \cos\theta = y \), \( x^2 + y^2 = 1 \) So, \( 2xy = -\frac{9}{14} \implies xy = -\frac{9}{28} \) Let’s solve: \[ x^2 + y^2 = 1 \\ xy = -\frac{9}{28} \] These are roots of quadratic: \[ t^2 - st - \frac{9}{28} = 0 \] where \( s = x+y \), \( p = xy \). But a quicker way: Use \( x^2 + y^2 = 1 \) Let’s use \( x = \sin\theta \), \( y = \cos\theta \) But to get the values, you can solve the system, but the above gives the method. --- ### 25. Verify the following identity: \[ \csc^2 t + \sec^2 t = \frac{\sec^2 t}{\sin^2 t} \] **Recall:** - \( \csc^2 t = \frac{1}{\sin^2 t} \) - \( \sec^2 t = \frac{1}{\cos^2 t} \) LHS: \[ \csc^2 t + \sec^2 t = \frac{1}{\sin^2 t} + \frac{1}{\cos^2 t} = \frac{\cos^2 t + \sin^2 t}{\sin^2 t \cos^2 t} = \frac{1}{\sin^2 t \cos^2 t} \] RHS: \[ \frac{\sec^2 t}{\sin^2 t} = \frac{1/\cos^2 t}{\sin^2 t} = \frac{1}{\sin^2 t \cos^2 t} \] Thus, **both sides are equal.** --- If you want a solution for a specific part in detail, let me know!