Questions and Answers #3 Right Triangles & Trigonometry

Questions and Answers Sheet 3 Right Triangles & Trigonometry Question #1 Find the values of x such that the angle between the vectors (2, 1, -1), and (1, x, 0) is 45∘ . Answer: Theorem 3 from this section states that: 𝑎 ⋅ 𝑏 = |𝑎||𝑏| cos 𝜃. where \theta is the angle between the vectors a,b ≪ 2,1, −1 ≫⋅≪ 1, 𝑥, 0 ≫ = √22 + 12 + (−1)2 √12 + 𝑥 2 + 0 cos 4 5∘ Remember that: ≪ 𝑎1 , 𝑏1 , 𝑐1 ≫⋅≪ 𝑎2 , 𝑏2 , 𝑐2 ≫= 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2 This is because when it comes to the unit vectors i,j,k, their dot product with themselves is 1 and their dot product with the other two is 0. 1 (2)(1) + (1)(𝑥) + (−1)(0) = √4 + 1 + 1√1 + 𝑥 2 + 0 √2 2 + 𝑥 + 0 = √6√1 + 𝑥 2 1 √2 2 + 𝑥 = √3√1 + 𝑥 2 Square both sides (2 + 𝑥)2 = 3(1 + 𝑥 2 ) 4 + 4𝑥 + 𝑥 2 = 3 + 3𝑥 2 Rewrite the quadratic equation in the standard form 2𝑥 2 − 4𝑥 − 1 = 0 Using the quadratic formula, we can write 𝑥= −(−4)±√(−4)2 −4(2)(−1) 2(2) = 4±√24 2(2) = 2±√6 2 Question #2 Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A (1, 0,-1), B (4, -3,0), C (1, 5 ,5) ∠𝐶𝐴𝐵 =____ ∘ ∠𝐴𝐵𝐶 =____ ∘ ∠𝐵𝐶𝐹 =____ ∘ ∠𝐶𝐴𝐵 = ? ∠𝐴𝐵𝐶 = ? ∠𝐵𝐶𝐴 = ? Answer: Step 1 Find there angles. Given vertices: А (1, 0, -1), В (4, -3, 0), С (1, 5, 5) . Distance formula: 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2 Step 2 a be the distance between B and C. 𝑎 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 02 ) 2 = √(1 − 4)2 + (5 − (−3)) + (5 − 0)2 = √(−3)2 + (8)2 + (5)2 = √(−3)2 + (8)2 + (5)2 7√2 Step 3 b be the distance between A and C. 𝑏 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧12 ) 2 = √(1 − 1)2 + (5 − 0)2 + (5 − (−1)) {= √(0)2 + (5)2 + (6)2 = √61 Step 4 c be the distance between A and B. 𝑐 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧12 ) = √(4 − 1)2 + (−3 − 0)2 + (0 − (−1)) 2 = √(3)2 + (−3)2 + (1)2 = √19 Step 5 Apply cosine law. To get an angle A. 𝑏 2 +𝑐 2 −𝑎2 cos 𝐴 = 2𝑏𝑐 2 = 2 (√61) +(√19) −(7√2) 2 2⋅√61⋅√19 =-0.2644 𝐴 = cos −1 (−0.2644) 𝐴 = 105.33∘ Question #3 (a) Write cos 3 𝑥 + 𝑖 sin 3 𝑥 = 𝑒 3𝑖𝑥 = (cos 𝑥 + 𝑖 sin 𝑥)3 by Euler's formula, expand, and equate real and imag inary parts to derive the identities 3 1 𝑐𝑜𝑠 3 𝑥 = cos 𝑥 + cos 3 𝑥 4 3 4 1 4 4 sin3 𝑥 = sin 𝑥 − sin 3 𝑥 (b) Use the result of part (a) to find a general solution of 𝑦 ′′ + 4𝑦 = cos 3 𝑥. Answer: a) Given The equation is cos 3 𝑥 + 𝑖 sin 3 𝑥 = 𝑒 3 𝑖𝑥 = (cos 3 𝑥 + 𝑖 sin 3 𝑥)3 Expand the equation cos 3 𝑥 + 𝑖 sin 3 𝑥 = cos 3 𝑥 + 3𝑖 cos 2 𝑥 sin 𝑥 − 3 cos 𝑥 sin2 𝑥 − 𝑖 sin3 𝑥 = cos 3 𝑥 + 3𝑖(1 − sin2 𝑥) sin 𝑥 − 3 cos 𝑥 (1 − cos 2 𝑥) − 𝑖 sin3 𝑥 = cos 3 𝑥 + 3𝑖 sin 𝑥 − 3𝑖 sin3 𝑥 − 3 cos 𝑥 + 3 cos 3 𝑥 − 𝑖 sin3 𝑥 = 4 cos 3 𝑥 + 3𝑖 sin 𝑥 − 3 cos 𝑥 − 4𝑖 sin3 𝑥 = (4 cos 3 𝑥 − 3 cos 𝑥) + 𝑖(3 sin 𝑥 − 4 sin3 𝑥) Equate the real part: cos 3 𝑥 = 4 cos 3 𝑥 − 3 cos 𝑥 4 cos 3 𝑥 = cos 3 𝑥 + 3 cos 𝑥 1 3 4 4 cos 3 𝑥 = cos 3 𝑥 + cos 𝑥 Equate the imaginary part: sin 3 𝑥 = 3 sin 𝑥 − 4 sin3 𝑥 4 sin3 𝑥 = 3 sin 𝑥 − sin 3 𝑥 3 1 sin3 𝑥 = sin 𝑥 − sin 3 𝑥 4 4 Hence proved Question #4 15 Given: sin 𝜃 = and 17 𝜋 2 < 𝜃 < 𝜋. Find the exact value of: cos A) 5√34 B) − C) 34 Montreal, Canada Barcelona, Spain 34 3√34 D) − 2 New York City 34 5√34 34 3√34 𝜃 Nassau, Bahamas Answer: 15 Ans: Given, sin 𝜃 = 17 𝜋 < 𝜃 < 𝜋. 2 We have to find exact value of cos Since, sin 𝜃 = 𝜃 2 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ⇒ Perpendicular = 15 Hypotenuse = 17 We know that, Pythagorean theorem 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2 = 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 2 + 𝐵𝑎𝑠𝑒 2 ⇒ (17)2 = (15)2 + 𝐵𝑎𝑠𝑒 2 ⇒ 𝐵𝑎𝑠𝑒 2 = 289 − 225 = 64 ⇒ 𝐵𝑎𝑠𝑒 = 8 𝐵𝑎𝑠𝑒 Since, cos 𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 8 ⇒ cos 𝜃 = 17 𝜋 ∵ < 𝜃 < 𝜋(2𝑛 𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡) 2 So, cos 𝜃 = −8 17 Now, We know that cos 2 𝜃 = 2 cos 2 𝜃 − 1 ⇒ cos 2 𝜃 = 𝜃 ⇒ cos 2 = 1+cos2 2 ⇒ cos ∵ 𝜋 2 𝜃 2 𝜃 2 1+cos 𝜃 =√ 2 1+cos 𝜃 <𝜃<𝜋⇒ 2 𝜋 4 < 𝜃 2 < 𝜋 2 (1𝑠𝑡 quadrant) −8 𝜃 √1 + ( 17 ) 17 − 8 9 3 3 √34 3√34 ⇒ cos = =√ =√ = = × = 2 2 17 × 2 34 √34 √34 √34 34 Hence, cos 𝜃 2 = 3√34 (34) ⇒ Barselona, Spain Question #5 Use half-angle identities to find an exact value without a calculator. cos 7 5∘ Answer: Let 75∘ = 150∘ 2 ∘ cos 7 5 = cos : 150∘ 2 Use the half-angle identity for sin 𝑢 2 = ±√ 1+cos 𝑢 2 where 𝑢 = 150∘ . Since 75∘ is in Ql, then we take the positive identity: =√ 1+cos 150∘ 2 By special angles, cos 1 50∘ = − =√ √3 2 √3 1+(− 2 ) 2 2 √3 −2 = √2 =√ =√ = 2 2−√3 2 2 2−√3 4 √2−√3 2 Result: √2−√3 2 Question #6 Rewrite each angle in radian measure as a multiple of 𝜋 (Do not use a calculator.) 120∘ Answer: Multiply by 𝜋/180 to get radians. 𝜋 2 120∘ × = 𝜋. 180∘ 3 Result: 2 𝜋. 3 Question #7 Find a*b. |a|=6, |b|=5, the angle between a and b is 2𝜋/3 Answer: The dot product can be found using the formula |𝑎| ∗ |𝑏| ∗ cos(0), where 0 is the smallest angle between a and b. 2𝜋 (6) ∗ (5) ∗ (cos ( )) 3 Multiply. 6 ∗ 5 ∗ (−.5) = −15 Result: -15 Question #8 Find the value of cos 2 40°. Answer: Reference angle for 240° is 60°: 240°=180°+60° 1 cos 6 0° = 2 As cos 2 40° = − cos 6 0°, we have cos 2 40° = − 1 2 Question #9 Why is cos(−𝑝𝑖/6) the same as cos(𝑝𝑖/6) ? Answer: The cosine function is even. This means that cos(−𝑝𝑖/6) is the same thing as cos(𝑝𝑖/6). Question #10 In a 30-60-90 triangle, the shorter leg has length of 8√3𝑚. What is the length of the other leg and the hypotenuse? Answer: Let the shortest length is represented by the variable a. The other lengths are multiples of a. If 𝑎 = 8√3 𝑎√3 = (8√3)√3 = 24 2𝑎 = 2(8√3) = 16√3 Therefore, the height - 24 , hypotenuse - 16√3