Topic 3 Geometry and Trigonometry

Topic 3: Geometry and Trigonometry PART A Distance, 3D Volume, Surface Area: Distance between 2 points in 3-Dimensional Space: 𝐷 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2 Midpoint of 2 points in 3-Dimensional Space: 𝑥2 + 𝑥1 𝑦2 + 𝑦1 𝑧2 + 𝑧1 𝑀=( , , ) 2 2 2 Volume and Surface Area of 3-D Solids: Square Pyramid: 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 𝑏 2 + 2𝑏𝑠 1 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑏 2 ℎ 3 Right Circular Cone: 1 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 2 ℎ 3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 𝜋𝑟 2 + 𝜋𝑟𝑙 Sphere: 4 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 3 3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 4𝜋𝑟 2 Hemisphere: 2 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 3 3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 2𝜋𝑟 2 IB Questions: 50 51 52 53 PART B Right-Angled and Non Right-Angled Trigonometry: 𝑠𝑖𝑛𝜃 = 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑛𝜃 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 , 𝑐𝑜𝑠𝜃 = , 𝑡𝑎𝑛𝜃 = = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑐𝑜𝑠𝜃 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 Sine Rule: Cosine Rule: Or Area of a triangle using sine: 1 𝐴𝑟𝑒𝑎 = 𝑎𝑏𝑠𝑖𝑛𝐶 2 Exercise 3B: Find X: 54 1. 2. 3. PART C 55 Applications of Right and Non Right-Angled Trigonometry: Now, we will look at the applications of Right and Non Right-Angled Trigonometry. The applications include bearings, angle of elevation and angle of depression. Exercise: 1. A flagpole is held up by 3 wires as given in the diagram below. a. Calculate angle ADB b. Determine the area of 𝛥𝐴𝐷𝐵 c. Determine the angle of elevation of the peak of the tower from point C d. Prove that 𝛥𝐴𝐵𝐶 is not a right-angled triangle. 2. Quadrilateral ABCD is given below. a. Find the length of AC b. Find angle BAC c. Calculate the size of ACD d. Find the length of AD e. Determine the area of the quadrilateral ABCD. 3. A ship leaves Jeju with a bearing of 030° and travels a distance of 250 km to Shanghai. At point B, the ship shifts direction with a bearing of 130°. It then travels a distance of 434 km to reach Nagasaki. Determine the bearing and distance for the ship to return from Nagasaki to Jeju. 56 PART D Radian measure of angles, Length of an arc, and Area of a sector: Radians are an alternative way of measuring angles. One radian is equal to the angle that is formed when an arc opposite the angle has the same length as the radius of a circle. It is illustrated in a more simple manner below: Converting between radians and degrees: 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 180 𝐷𝑒𝑔𝑟𝑒𝑒𝑠 Length of an arc: In degrees, the length of an arc is equal to: 𝑙= 𝜃 × 2𝜋𝑟 360 Converting to radians: 𝑙 = 𝜃𝑟 Area of a sector: In degrees, the area of a sector is equal to: 𝜃 × 𝜋𝑟 2 360 Converting to radians: 𝐴 = 𝜃𝑟 2 𝐴= Exercise: 1. Find the length of the arc and the area of the sector: 57 a. b. c. IB Questions: 58 PART E Unit Circle: The equation of a unit circle is as follows: 𝑥 2 + 𝑦 2 = 1. 59 𝑥 = 𝑟𝑐𝑜𝑠𝜃 𝑦 = 𝑟𝑠𝑖𝑛𝜃 r =1 𝐻𝑒𝑛𝑐𝑒, 𝑥 = 𝑐𝑜𝑠𝜃 𝑎𝑛𝑑 𝑦 = 𝑠𝑖𝑛𝜃 Thus, we can deduce (𝑐𝑜𝑠𝜃) + (𝑠𝑖𝑛𝜃)2 = 1 Note: −1 ≤ 𝑥 ≤ 1 𝑎𝑛𝑑 − 1 ≤ 𝑦 ≤ 1. Thus, we can conclude that −1 ≤ 𝑐𝑜𝑠𝜃 ≤ 1 and −1 ≤ 𝑠𝑖𝑛𝜃 ≤ 1. This will be very important when we graph these functions. 2 60 Now, we can understand where a trigonometric function is positive. A point in the first quadrant is given by (x, y), thus, both 𝑐𝑜𝑠𝜃and 𝑠𝑖𝑛𝜃 will be positive and as, 𝑠𝑖𝑛𝜃 tan𝜃is 𝑐𝑜𝑠𝜃, it too is positive. A point in the second quadrant is given by (-x, y), thus, while 𝑠𝑖𝑛𝜃 is positive, both 𝑐𝑜𝑠𝜃 and tan𝜃are negative. A point in the third quadrant is given by (-x, -y), thus, both 𝑐𝑜𝑠𝜃and 𝑠𝑖𝑛𝜃 will be negative and 𝑠𝑖𝑛𝜃 as, tan𝜃is 𝑐𝑜𝑠𝜃, it is positive. A point in the second quadrant is given by (x, -y), thus, while 𝑐𝑜𝑠𝜃 is positive, both 𝑠𝑖𝑛𝜃 and tan𝜃are negative. This concept can be remembered by a simple phrase: All Silver Tea Cups. 61 We need to know the values of the unit circle. The diagram below illustrates all the important ones. While this may seem confusing, there is an easier method. We only need to learn the sine and 𝑠𝑖𝑛𝜃 cosine values in the first quadrant and use ‘All Silver Tea Cups’ and the relation tan𝜃= 𝑐𝑜𝑠𝜃to get everything else. Necessary sine and cosine values: ● 𝑐𝑜𝑠 0 = 𝑠𝑖𝑛 90 = 𝑠𝑖𝑛 2 = 1 𝜋 𝜋 ● 𝑠𝑖𝑛 0 = 𝑐𝑜𝑠 90 = 𝑐𝑜𝑠 2 = 0 𝜋 1 𝜋 √3 . 2 𝜋 ● 𝑠𝑖𝑛 30 = 𝑠𝑖𝑛 6 = 2. This is also equal to cos60 and cos 3. ● 𝑐𝑜𝑠 30 = 𝑐𝑜𝑠 6 = 𝜋 This is also equal to sin60 and sin3. 𝜋 𝜋 ● 𝑠𝑖𝑛 45 = 𝑠𝑖𝑛 4 = 𝑐𝑜𝑠 45 = 𝑐𝑜𝑠 4 = √2 2 . Other important relations: 1 1. 𝑠𝑒𝑐𝜃 = 𝑐𝑜𝑠𝜃 1 2. 𝑐𝑜𝑠𝑒𝑐𝜃 = 𝑠𝑖𝑛𝜃 1 3. 𝑐𝑜𝑡𝜃 = 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 Trigonometric Identities: Supplementary Angles: 1. 𝑐𝑜𝑠(180 − 𝜃) = 𝑐𝑜𝑠(𝜋 − 𝜃) = −𝑐𝑜𝑠𝜃 2. 𝑠𝑖𝑛(180 − 𝜃) = 𝑠𝑖𝑛(𝜋 − 𝜃) = 𝑠𝑖𝑛𝜃 3. 𝑡𝑎𝑛(180 − 𝜃) = 𝑡𝑎𝑛(𝜋 − 𝜃) = −𝑡𝑎𝑛𝜃 Negative Angles: 1. 𝑐𝑜𝑠(−𝜃) = 𝑐𝑜𝑠𝜃 2. 𝑠𝑖𝑛(−𝜃) = 𝑠𝑖𝑛𝜃 3. 𝑡𝑎𝑛(−𝜃) = −𝑡𝑎𝑛𝜃 Complementary Angles: 𝜋 1. 𝑐𝑜𝑠(90 − 𝜃) = 𝑐𝑜𝑠( 2 − 𝜃) = 𝑠𝑖𝑛𝜃 𝜋 2. 𝑠𝑖𝑛(90 − 𝜃) = 𝑠𝑖𝑛( 2 − 𝜃) = 𝑐𝑜𝑠𝜃 𝜋 1 3. 𝑡𝑎𝑛(90 − 𝜃) = 𝑡𝑎𝑛( 2 − 𝜃) = 𝑡𝑎𝑛𝜃 These identities are very important for simplifying various sums in the IB Syllabus and we will solve some of them together. 62 Example: Evaluate 2𝜋 (𝑠𝑖𝑛 3 )2 − 7𝜋 (𝑡𝑎𝑛 4 )2 . We know that 𝑠𝑖𝑛(𝜋 − 𝜃) = 𝑠𝑖𝑛𝜃. Thus 𝑠𝑖𝑛 𝜋 We can easily deduce that 𝑡𝑎𝑛 4 = 1. that quadrant. Thus, 𝑡𝑎𝑛 7𝜋 4 7𝜋 4 2𝜋 3 = 𝑠𝑖𝑛(𝜋 − 2𝜋 3 𝜋 ) = 𝑠𝑖𝑛 3 = √3 . 2 3 Thus, its square is 4. lies in the 4th quadrant and the tan function is negative in = −1. Hence, its square is 1. Therefore, the above expression simplifies to: 3 4 −1= −1 4 . Note- The equation of a straight line through the origin is y = xtanθ, where θ is the angle formed between the line and positive x-axis. Exercise 3E: 1. 2. Solve: a. 4(𝑠𝑖𝑛𝑥)2 − 3 = 0, 0 ≤ 𝑥 ≤ 2𝜋 b. 2(𝑡𝑎𝑛𝑥)2 -5=0, -2𝜋 ≤ 𝑥 ≤ 2𝜋 3 3 3. If 𝑠𝑖𝑛𝑥 = − 5 , 𝜋 ≤ 𝑥 ≤ 2 𝜋, find (hint, draw a right angled triangle to help yourself out) a. b. c. d. e. f. g. h. cosx tanx cos(𝜋-x) tan(2𝜋-x) cotx secx cosecx 𝜋 cosec(2-x) 𝜋 i. sec(2+x) 63 PART F Trigonometric Identities and Double Angle Formula: Important Trigonometric Identities: 1. (𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃)2 = 1 2. (𝑡𝑎𝑛𝜃)2 + 1 = (𝑠𝑒𝑐𝜃)2 3. (𝑐𝑜𝑡𝜃)2 + 1 = (𝑐𝑜𝑠𝑒𝑐𝜃)2 Double Angle Formulae: 1. 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 2. 𝑐𝑜𝑠2𝑥 = (𝑐𝑜𝑠𝑥)2 − (𝑠𝑖𝑛𝑥)2 = 2(𝑐𝑜𝑠𝑥)2 − 1 = 1 − 2(𝑠𝑖𝑛𝑥)2 2𝑡𝑎𝑛𝑥 3. 𝑡𝑎𝑛2𝑥 = 1−(𝑡𝑎𝑛𝑥)2 Exercise: 1. If x is acute and 𝑐𝑜𝑠2𝑥 = −7 9 , find a. Cosx b. Sinx c. Tanx 21 2. Find the exact value of tanA if 𝑡𝑎𝑛2𝐴 = 20and A is obtuse. 3. Prove that: a. 𝑠𝑖𝑛2𝑥 1−𝑐𝑜𝑠2𝑥 = 𝑐𝑜𝑡𝑥 b. (𝑐𝑜𝑠𝜃)4 − (𝑠𝑖𝑛𝜃)4 = 𝑐𝑜𝑠2𝜃 c. 𝑐𝑜𝑠𝑒𝑐2𝜃 = 𝑡𝑎𝑛𝜃 + 𝑐𝑜𝑡2𝜃 d. 𝑠𝑖𝑛2𝐴 𝑠𝑖𝑛𝐴 − 𝑐𝑜𝑠2𝐴 𝑐𝑜𝑠𝐴 = 𝑠𝑒𝑐𝐴 4. Solve for x, 0 ≤ 𝑥 ≤ 2𝜋 a. 𝑠𝑖𝑛2𝑥 − 𝑐𝑜𝑠𝑥 = 0 b. 𝑐𝑜𝑠2𝑥 − 𝑠𝑖𝑛𝑥 = 0 c. 𝑐𝑜𝑠4𝑥 = 𝑐𝑜𝑠2𝑥 d. 2(𝑠𝑖𝑛𝑥)2 = 3𝑐𝑜𝑠𝑥 e. 3(𝑡𝑎𝑛𝑥)2 − 13𝑡𝑎𝑛𝑥 + 4 = 0. (𝐻𝑖𝑛𝑡, 𝑡ℎ𝑖𝑛𝑘 𝑏𝑎𝑐𝑘 𝑡𝑜 22𝑥 𝑡𝑦𝑝𝑒 𝑠𝑢𝑚𝑠. ) PART G Graphing and Modelling Trigonometric Functions: Graphing Trigonometric Functions: The graph of sin(x) is given below for 0 ≤ 𝑥 ≤ 2𝜋: 64 As sin(x) is a periodic function, the function exists in a similar manner for 𝑥𝜖𝑅. The graph of cos(x) is given below for 0 ≤ 𝑥 ≤ 2𝜋 As cos(x) is a periodic function, the function exists in a similar manner for 𝑥𝜖𝑅. 𝑠𝑖𝑛𝜃 However, tan𝜃= 𝑐𝑜𝑠𝜃. Thus, the function has a vertical asymptote (discussed in topic 2) whenever 𝜋 𝜋 3𝜋 cos𝜃is equal to zero, which is at 𝑥 = 2 , − 2 , 𝜋 at 𝑥 = 2 (2𝑛 + 1), 𝑤ℎ𝑒𝑟𝑒 𝑛𝜖𝑍. 2 ,− 3𝜋 2 , . ..Essentially, there are vertical asymptotes 𝜋 Thus, the graph of tan𝜃is given by the following for0 ≤ 𝑥 ≤ 2𝜋, 𝑥 ≠ 2 (2𝑛 + 1) : 65 The equation for a general sine function is given by the following: f(x) = asin(b(x + c)) + d, where a is the amplitude of the function, b is the period, c is the horizontal translation and d is the vertical translation. The amplitude is the difference between the mean position of the graph and its highest/lowest point. The period of a trigonometric function is the minimum time it takes for a function to return from its mean position after reaching a maximum and a minimum. The period of the most simple sine graph, y=sinx, is 2𝜋. Here, B=1, thus, we can conclude that 𝑃𝑒𝑟𝑖𝑜𝑑 = cosine function is 2𝜋 𝐵 2𝜋 𝐵 .Similar to a sine function, the period of a and its amplitude is the difference between the mean position of the graph and its highest/lowest point. Additionally, the range of basic sine and cosine functions is −1 ≤ 𝑦 ≤ 1. However, the range of tan function is 𝑦𝜖𝑅. Moreover, the period of the most simple tan 𝜋 function is 𝜋. Thus, its period in general is given by 𝐵.”à” for a tan function is not the amplitude, rather, it is the vertical stretch of the function. Note:- We can convert sine and cosine graphs to each other very easily. In the most basic sine 𝜋 function, it can be transformed to a cosine function by a horizontal translation by 2units to the left. Modelling Trigonometric Functions: Example: 66 1. Given below is a table that has recorded the average monthly temperatures for a city in India. a. Plot the given data on a graph, taking the months as x-values (For example, Jan=1, Feb=2 and so on) b. This data can be modelled by a trigonometric function in the form T= asin(b(x + c)) + d. Without using technology, estimate the values of i. b ii. a iii. d iv. c c. Check how well your model fits the data using technology. The first subpart is quite simple and just requires attention to detail while graphing. Next, roughly connect the points in a form resembling a trigonometric function and then analyse the graph to roughly estimate the period of the graph. To determine the amplitude a, we must look at the maximum and minimum temperature. 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 = translation is given by 𝑑 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚+𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2 𝑚𝑎𝑥𝑖𝑚𝑢𝑚−𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2 .The vertical . Finally, with only one variable left (c), we can simply use substitution to get its value. Exercise: 1. For the function 𝑦 = 3𝑠𝑖𝑛( 4 ) + 2, find the following: 𝑥 a. Amplitude b. Period c. Range 2. Sketch the following functions for 0 ≤ 𝑥 ≤ 3𝜋: 𝜋 a. 𝑦 = 2𝑐𝑜𝑠(𝑥 + 3 ) − 1 b. 𝑦 = 𝑠𝑖𝑛2𝑥 + 5 c. 𝑦 = 3𝑡𝑎𝑛(2𝑥 − 𝜋) − 2 3. State the transformations which transform: 𝜋 a. 𝑦 = 𝑠𝑖𝑛 𝑥 𝑡𝑜 𝑦 = 𝑠𝑖𝑛(2(𝑥 − 4 ) + 2 1 1 2 b. 𝑦 = 𝑡𝑎𝑛 𝑥 𝑡𝑜 𝑦 = 5 𝑡𝑎𝑛(2 (𝑥 + 𝜋)) + 5 4. On a June day in London, the maximum temperature is 18.7 °C occurs at 1:30 pm and the minimum temperature is 4.5°C. Suggest a cosine function that can model the temperature for that day, where T is the temperature and t is the hours after midnight. 67 PART H Reciprocal Trigonometric Functions and Inverse Trigonometric Functions: Thus, from the unit circle: 1. 𝑠𝑖𝑛𝜃 = 𝐵𝐹 2. 𝑐𝑜𝑠𝜃 = 𝑂𝐹 3. 𝑡𝑎𝑛𝜃 = 𝐴𝐶 4. 𝑠𝑒𝑐𝜃 = 𝑂𝐶 5. 𝑐𝑜𝑠𝑒𝑐𝜃 = 𝑂𝐺 6. 𝑐𝑜𝑡𝜃 = 𝐸𝐺 Sketching reciprocal functions: 𝜋 3𝜋 𝑦 = 𝑠𝑒𝑐𝜃for 0 ≤ 𝜃 ≤ 2𝜋: (It has vertical asymptotes at 𝑥 = 2 , 2 68 𝑦 = 𝑐𝑜𝑠𝑒𝑐𝜃 𝑓𝑜𝑟 0 ≤ 𝜃 ≤ 2𝜋: (It has vertical asymptotes at 𝑥 = 0, 𝜋, 2𝜋) We can easily deduce the graph of 𝑦 = 𝑐𝑜𝑡𝜃 𝑓𝑜𝑟 0 ≤ 𝜃 ≤ 2𝜋, using our learning from 𝑦 = 1 in advanced transformations in functions. 𝑓(𝑥) Sketching inverse trigonometric functions: 69 Let us first start with 𝑦 = 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 (𝑠𝑖𝑛−1 (𝑥) 𝑓𝑜𝑟 − 1 ≤ 𝑥 ≤ 1 Domain𝜖[−1,1] 𝜋 𝜋 Range𝜖 [− 2 , 2 ] Now, let us move onto 𝑦 = 𝑎𝑟𝑐𝑐𝑜𝑠𝑥, 𝑖𝑒, (𝑐𝑜𝑠 −1 (𝑥) 𝑓𝑜𝑟 − 1 ≤ 𝑥 ≤ 1 Domain 𝜖[−1,1] Range𝜖[0, 𝜋] Finally, we will look at the graph of 𝑦 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥, 𝑖𝑒, (𝑡𝑎𝑛−1 (𝑥) Domain: 𝐷𝜖𝑅 𝜋 𝜋 Range: 𝑅𝜖]− 2 , 2 [ 70 Exercise: 1. Simplify: 𝜋 a. arccos(cos 3) 𝜋 b. arcsin(sin− 4 ) 4𝜋 c. arctan(tan 3 ) 2. Find the other 5 trigonometric ratios if 2 3𝜋 a. 𝑠𝑖𝑛(𝑥) = − 3 , 2 5 < 𝑥 < 2𝜋 b. 𝑡𝑎𝑛(𝑥) = 3 , 𝜋 < 𝑥 < 3𝜋 2 3 𝜋 c. 𝑠𝑒𝑐(𝑥) = − 1 , 2 < 𝑥 < 𝜋 PART I Compound Angle Identities: Exercise: 5 1 1. Determine exactly the tangent of the acute angle between 2 lines with gradients 3and 7 2. If √2𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 = 𝑘𝑠𝑖𝑛(𝑥 − 𝑎), 𝑓𝑜𝑟 𝑘 > 0 𝑎𝑛𝑑 0 < 𝑎 < 2𝜋 . 3. We know that sinB=cosA+cosC. Prove that triangle ABC is right-angled. 6 3 4. If 𝑡𝑎𝑛(𝐴 − 𝐵) = 7and 𝑡𝑎𝑛𝐵 = 4, find the exact value of tanA. IB Questions: 𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴 1. Prove that 𝑐𝑜𝑠𝐴−𝑠𝑖𝑛𝐴 = 𝑠𝑒𝑐2𝐴 + 𝑡𝑎𝑛2𝐴 2. In triangle ABC, 3𝑠𝑖𝑛𝐵 + 4𝑐𝑜𝑠𝐶 = 6and 4𝑠𝑖𝑛𝐶 + 3𝑐𝑜𝑠𝐵 = 1 1 a. Prove that 𝑠𝑖𝑛(𝐵 + 𝐶) = 2 b. Show that angle CAB can have only one value 3. The diagram below illustrates the triangle ABC where AB=2, 𝐴𝐶 = √2,and angle BAC=15° 71 a. Expand and simplify (1 − √3)2 b. By writing 15° as 60°-45°, determine the value of cos(15°) c. Determine the value of BC in the form 𝑎 + √𝑏 , where a, b ∈ 𝑍 𝜋 𝜋 4. The first 3 terms of a geometric sequence are sinx, sin2x, 4sinx(cosx)2, − 2 < 𝑥 < 2 . a. Find the common ratio r b. Find the set of values of x for which geometric series sinx+sin2x+ 4sinx(cosx)2+.. converges 1 c. Let 𝑥 = 𝑎𝑟𝑐𝑐𝑜𝑠 (4) , 𝑥 > 0. Show that the sum to infinity of this series is √15 . 2 5. Obtain all the solutions for the equation 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛2𝑥 = 0 𝑓𝑜𝑟 0 < 𝑥 < 360° 3 𝜋 6. Given that 2 < 𝑥 < 𝜋 and 𝑐𝑜𝑠𝑥 = − 4,find the value of sin2x. 7. In triangle ABC, angle ABC=90°, 𝐴𝐶 = √2, 𝐴𝐵 = 𝐵𝐶 + 1. a. Prove that 𝑐𝑜𝑠𝐴 − 𝑠𝑖𝑛𝐴 = 1 √2 b. By squaring both sides of the equation in part a, solve the equation to find the angles in the triangle c. Use Pythagoras theorem in triangle ABC to get BC, and hence, prove that 𝑠𝑖𝑛𝐴 = √6−√2 4 d. Hence, or otherwise, calculate the length of the perpendicular from B to [AC] 8. (With Calculator) a. Solve the equation 3(𝑐𝑜𝑠𝑥)2 − 8𝑐𝑜𝑠𝑥 + 4 = 0, 𝑤ℎ𝑒𝑟𝑒0 < 𝑥 < 180°, expressing your answer to the nearest degree b. Determine the exact value of sec x that satisfy the equation 3(𝑠𝑒𝑐𝑥)4 − 8(𝑠𝑒𝑐𝑥)2 + 4 = 0, 9. (Without Calculator) 𝜋 a. Sketch the graphs of 𝑦 = 𝑐𝑜𝑠𝑥and 𝑦 = 3𝑐𝑜𝑠2𝑥 − 1 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 0 ≤ 𝑥 ≤ 2 . b. Find the x-coordinates of the point of intersection of the two graphs for the same domain given above. 72