PART 2: Investigate and rigorously prove the identity: \( \sin ^{2} A+\cos ^{2} A=1 \) using coordinate geometry and algebraic manipulation. Prior knowledge required: - Definition of sine and cosine in a right-angled triangle. - Pythagoras Theorem. - Cartesian plan representation of trigonometric functions. TASK 1: Using \( \triangle \) AOP drawn in Part 1. Marks: 12 Let \( x \)-coordinate of P be \( \cos A \) and the \( y \)-coordinate of P be \( \sin A \) respectively. a) Express \( \triangle \mathrm{AOP} \) as Pythagoras Theorem in terms of \( \sin A \) and \( \cos A \). (3) b) Provide numerical verification by using THREE acute special angles (e.g., \( A=30^{\circ} \) ) and substitute into the equation formulated above in a). Show all your workings. c) What mathematical deduction can be made from b) above? (2) PART 3: Investigating tangent identity. Marks: 11 TASK 1: 1. Draw any right-angled triangle \( A B C \) with: (1) 2. Write down the definitions of the following trigonometric ratios using the triangle drawn: a) \( \sin A \) b) \( \cos A \) c) \( \quad \tan A \) 3. Express \( \frac{\sin A}{\cos A} \) using the definitions above, in terms of \( x \) and \( y \), and simplify to the simplest fractional form. 4. Compare the answer obtained above in question 3 and ratio obtain in 2.c). 5. Verify the validity of your answer in 4.) by calculations attaching values to Hypotenuse, Opposite and Adjacent. Please turn over Copyright reserved

To rigorously prove the identity \( \sin^2 A + \cos^2 A = 1 \) using coordinate geometry and algebraic manipulation, we will follow the steps outlined in the problem. ### PART 2: #### TASK 1: **a)** Express \( \triangle \mathrm{AOP} \) as Pythagorean Theorem in terms of \( \sin A \) and \( \cos A \). In a right triangle \( \triangle AOP \), where: - \( O \) is the origin (0, 0), - \( A \) is the point on the x-axis (1, 0), - \( P \) is the point on the unit circle corresponding to angle \( A \) with coordinates \( (\cos A, \sin A) \). According to the Pythagorean Theorem, the relationship between the sides of the triangle is given by: \[ OP^2 = OA^2 + AP^2 \] Here: - \( OP \) is the hypotenuse (which is 1, since \( P \) lies on the unit circle), - \( OA \) is the adjacent side (which is \( \cos A \)), - \( AP \) is the opposite side (which is \( \sin A \)). Thus, we can write: \[ 1^2 = \cos^2 A + \sin^2 A \] This simplifies to: \[ \sin^2 A + \cos^2 A = 1 \] **b)** Provide numerical verification by using THREE acute special angles (e.g., \( A=30^{\circ} \)) and substitute into the equation formulated above in a). Let's verify the identity for three special angles: \( A = 30^{\circ}, 45^{\circ}, 60^{\circ} \). 1. For \( A = 30^{\circ} \): - \( \sin 30^{\circ} = \frac{1}{2} \) - \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) - Calculation: \[ \sin^2 30^{\circ} + \cos^2 30^{\circ} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \] 2. For \( A = 45^{\circ} \): - \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) - Calculation: \[ \sin^2 45^{\circ} + \cos^2 45^{\circ} = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} + \frac{2}{4} = 1 \] 3. For \( A = 60^{\circ} \): - \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) - \( \cos 60^{\circ} = \frac{1}{2} \) - Calculation: \[ \sin^2 60^{\circ} + \cos^2 60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1 \] **c)** What mathematical deduction can be made from b) above? From the calculations above, we can deduce that for the angles \( 30^{\circ}, 45^{\circ}, \) and \( 60^{\circ} \), the identity \( \sin^2 A + \cos^2 A = 1 \) holds true. This confirms that the identity is valid for these acute angles, supporting its general applicability in trigonometry. ### PART 3: Investigating tangent identity. #### TASK 1: 1. Draw any right-angled triangle \( ABC \) with: - \( A \) as the angle of interest, - \( B \) as the right angle, - \( C \) as the opposite vertex. 2. Write down the definitions of the following trigonometric ratios using the triangle drawn: - a) \( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} \) - b) \( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \) - c) \( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \) 3. Express \( \frac{\sin A}{\cos A} \) using the definitions above, in terms of \( x \) and \( y \), and simplify to the simplest fractional form. Using the definitions: \[ \frac{\sin A}{\cos A} = \frac{\frac{BC}{AC}}{\frac{AB}{AC}} = \frac{BC}{AB} = \tan A \] 4. Compare the answer obtained above in question 3 and the ratio obtained in 2.c). From question 3, we have \( \frac{\sin A}{\cos A} = \tan A \), which is consistent with the definition of tangent in question 2.c). 5. Verify the validity of your answer in 4.) by calculations attaching values to Hypotenuse, Opposite, and Adjacent. Let’s assign values: - Hypotenuse \( AC = 5 \) - Opposite \( BC = 3 \) - Adjacent \( AB = 4 \) Calculating: - \( \sin A = \frac{BC}{AC} = \frac{3}{5} \) - \( \cos A = \frac{AB}{AC} = \frac{4}{5} \) Now, calculate \( \tan A \): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \] This matches the ratio \( \frac{BC}{AB} = \frac{3}{4} \), confirming the validity of the tangent identity. Thus, we have rigorously proven the identity \( \sin^2 A + \cos^2 A = 1 \) and verified the tangent identity through geometric and algebraic methods.