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Math 1613 - Trigonometry Homework #6 - 2007.09.27 Due Date - 2007.10.04 Solutions 1. Find the reference angle θR if θ has the given measure. Also, state which quadrant each of the original angles lies in. a) 143π 4 3 So notice that 143π 4 = 35π + 4 π. The angle 35π gets us to the negative x-axis, and rotating another 43 π gives that the angle lies in the fourth quadrant and θR = 14 π. b) 840◦ Here we have 840◦ = 720◦ + 120◦. So our angle lies in the second quadrant with θR = 180◦ − 120◦ = 60◦ = π3 . c) − 43π 3 First, − 43π = −14π − 31 π. This gives an angle in the fourth quadrant with 3 θR = π3 π. 2. Find the exact value. a) cos − 43π 3 Since cosine keeps the same sign in the fourth quadrant, we have that π 1 43π cos − = cos = . 3 3 2 b) csc 143π 4 Sine changes sign in the fourth quadrant when compared to the first quadrant, so we have π √ 143π csc = − csc = − 2. 4 4 1 c) sec (840◦) Cosine is negative in the second quadrant, so π ◦ sec (840 ) = − sec = −2. 3 3. For this problem, please refer to the following figure. 10 y θ x a) If sin(θ) = 12 , what is y? Since sin(θ) = 12 , we look at the picture and see that 1 y sin(θ) = = → y = 5. 2 10 b) If cos(θ) = 15 , what is sin(θ)? (Solve this by the Pythagorean Theorem and the definition of sin(θ)). If cos(θ) = 15 , then similar to part a), we have that x = 2. Next, we need to find y, which can be√done by√the Pythagorean Theorem. So√102 = 42 + y 2 and solving √ for y gives y = 96 = 4 6. So finally, sin(θ) = 4106 = 2 5 6 . c) Redo part b) but this time, use the fact that sin2 (θ) + cos2(θ) = 1. Since we know that cos(θ) = 15 , we get s r √ 2 r 1 1 24 2 6 sin(θ) = 1 − = 1− = = . 5 25 25 5 2 d) If tan(θ) = 3, what are x and y? Since tan(θ) = 3, we have that 3 = Pythagorean Theorem, we get that y x which means that x = 13 y. Using the √ 1 2 2 x + y = 100 → y + y = 100 → y = 3 10. 9 2 2 Since we now know y, we can solve for x in x2 + y 2 = 100. This gives x = Notice that xy = 3. 3 √ 10.