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Exam 2 Math 1151, Precalculus 2 Name: Summer 2012 Instructions. 1. Show all the steps of your work clearly. 2. No graphing calculators, iPods, phones or computers Question Points Your Score Q1 30 Q2 25 Q3 15 Q4 10 Q5 20 Q6 20 Q7 20 Q8 20 Q9 20 TOTAL 180 Q1]. . . [30 points] Let f (x) = −2 sin(2πx + π) − 1. (a) What is the amplitude? 2 (b) What is the period T ? 1 (c) What is the phase shift φ ω ? −1/2 (d) Graph one cycle of f (x) on the interval [ ωφ , ωφ + T ] and label the five key points. We are going to graph one cycle on the interval [−1/2, 1/2] To get the five key points we need to divide [−1/2, 1/2] into four equal subintervals: [−1/2, −1/4]∪[−1/4, 0]∪[0, 1/4]∪[1/4, 1/2]. The endpoints give us the x-coordinates of the key points. So the key points are: (−1/2, −1), (−1/4, −3), (0, −1), (1/4, 1), (1/2, −1). Q2]. . . [25 points] Let f (x) = −4 sec(4x). (a) What is the period of the function? T = π/2 (b) What are the asymptotes in the interval (−π, π)? The asymptotes occur when cos(4x) = 0 and this happens when 4x = x= πk 2 π 2 + πk where k ∈ Z. Solving for + π8 . So the asymptotes in the interval (−π, π) are: ± π8 , ± 3π , ± 5π , ± 7π . 8 8 8 Q3]. . . [15 points] Find the exact value of cot(sec−1 (−3))? (DO NOT write the answer as a decimal) Let θ = sec−1 (−3) so that sec θ = −3 where 0 ≤ θ ≤ π but θ cannot be π/2. Then cos θ = − 31 , and thus √ √ √ p θ θ ∈ ( π2 , π]. A trigonometric identity gives us sin θ = 1 − 1/9 = 38 . Then cot θ = cos = − 88 = − 42 . sin θ Q4]. . . [10 points] What is the sum formula for cos(α + β)? cos(α + β) = cos α cos β − sin α sin β Q5]. . . [20 points] Given that tan α = 5 12 where π < α < 3π 2 and sin β = − 12 where π < β < 3π , 2 find cos(α + β) using the formula in Q4? p We need to find cos α, cos β and sin α. Using a trigonometric identity we find that cos β = − 1 − 1/4 = √ − 3 . 2 Now to find the other two trig functions we let y = −5 and x = −12 from tan α and the location of 5 the angle α. Then r = 13 and cos α = − 12 and sin α = − 13 . So cos(α + β) = 13 √ 12 3−5 . 26 Q6]. . . [20 points] Write the trigonometric expression cot(csc−1 u) as an algebraic expression in u. √ Let θ = csc−1 u so that csc θ = u where −π ≤ θ ≤ π2 but θ cannot be 0. Then cot θ = ± csc2 θ − 1 = 2 √ ± u2 − 1. Now we do not know whether to choose positive or the negative because cot is positive in the first quadrant but negative in the fourth quadrant. Q7]. . . [20 points] Prove the following identity by making the left side of the equation look like the right side. 1− 1− sin2 (θ) 1+cos(θ) = 1+cos θ−(1−cos2 θ) 1+cos θ = sin2 (θ) = cos(θ). 1 + cos(θ) cos θ(1+cos θ) 1+cos θ = cos θ. Q8]. . . [20 points] Find the inverse of the function f (x) = cos(x + 2) + 1. State the domain of f −1 . f −1 (x) = arccos(x − 1) − 2. Recall that the domain of the inverse cosine function is [−1, 1] and so −1 ≤ x − 1 ≤ 1 and so the domain of f −1 is 0 ≤ x ≤ 2 Q9]. . . [20 points] Solve the trigonometric equation cos2 θ − sin2 θ + sin θ = 0 on the interval [0, 2π]. Using the trigonometric identity cos2 θ = 1−sin2 θ we find that cos2 θ −sin2 θ +sin θ = −2 sin θ +sin θ +1 = −(2 sin θ + 1)(sin θ − 1) = 0. Therefore it must be that sin θ = 1 or sin θ = − 21 . The former has solution and the latter has solutions 11π 7π , 6. 6 π 2