5 Questions on Pre-Calculus with Solution - Midterm Exam 2 | Math 1, Exams of Pre-Calculus

Download 5 Questions on Pre-Calculus with Solution - Midterm Exam 2 | Math 1 and more Exams Pre-Calculus in PDF only on Docsity! MATH 1B, Lecture E (44059) Midterm Exam 2 [100 points] - Key, Sunday, November 21, 2004 Fall 2004, Dr. Masayoshi Kaneda, University of California, Irvine Name (Printed): Student ID: • You have 50 MINUTES to answer the following problems. • SHOW ALL YOUR WORK. NO PARTIAL CREDIT will be given for an answer without work except for #1. • SIMPLIFY YOUR ANSWERS AS MUCH AS POSSIBLE. • NO CALCULATOR of any type is allowed. 1. [10] Given y = −4 + 1 3 cos(1 2 x − 1). (1) [2] What is the amplitude? (2) [2] What is the vertical shift? (3) [3] What is the period? (4) [3] What is the phase shift when compared with y = −4+ 1 3 cos(1 2 x)? Solution: In general, if you rewrite to the form y = a sin(b(x−c))+d (a 6= 0, b 6= 0), the amplitude is |a|, the period is 2π|b| , the vertical shift is d, and the phase shift compared with y = a sin(bx) + d is c. Thus, in this problem, the amplitude is 1 3 , the vertical shift is −4 , the period is 2π 1/2 = 4π , and the phase shift compared with y = 1 − 2 sin(3x) is 2 . 2. [10] Evaluate. Solution: Note that the domain of the cosine function is restricted to [0, π]. (1) [2] cos−1(−1) = π since cos π = −1. (2) [2] cos(cos−1 0) = cos π 2 = 0 . (3) [2] cos−1(cos π 3 ) = cos−1 1 2 = π 3 since cos π 3 = 1 2 . (4) [4] cos−1(cos(−π 3 )) = cos−1 1 2 = π 3 . The inverse property does not hold since −π 3 is not on the restricted domain [0, π] of the cosine function. 1 2 3. [10] Let θ is in Quadrant II and cos θ = − 4 5 . (1) [4] Find sin θ. (2) [4] Find cos(2θ). (3) [2] Which quadrant is 2θ in? Solution: (1) sin2 θ = 1 − cos2 θ = 1 − ( −4 5 )2 = 1 − 16 25 = 9 25 . Since θ is in Quadrant II, sin θ > 0. Thus sin θ = √ 9 25 = 3 5 . Note: Remember cos2 θ + sin2 θ = 1. (2) cos(2θ) = cos2 θ − sin2 θ = 16 25 − 9 25 = 7 25 . Note: You may use that cos(2θ) = 2 cos2 θ − 1, or cos(2θ) = 1 − 2 sin2 θ too. These follow from cos(α + β) = cos α cos β − sinα sinβ by putting α = β = θ. Also remember that sin(2θ) = 2 sin θ cos θ. This follows from sin(α + β) = sinα cos β + cos α sinβ by putting α = β = θ. (3) Since θ is in Quadrant II, π 2 < θ < π, and hence, π < 2θ < 2π. But since cos(2θ) = 7 25 > 0, 3π 2 < 2θ < 2π, so that 2θ is in Quadrant IV . 4. [50] (1) [10] Find cos 75◦ by considering as cos(45◦ +30◦) and using the sum identity. Solution: cos 75◦ = cos(45◦+30◦) = cos 45◦ cos 30◦−sin 45◦ sin 30◦ = 1√ 2 √ 3 2 − 1√ 2 1 2 = √ 3 − 1 2 √ 2 . Note: Memorize sin(α + β) = sinα cos β + cos α sinβ, sin(α − β) = sinα cos β − cos α sinβ, cos(α + β) = cos α cosβ − sinα sinβ, cos(α − β) = cos α cosβ + sinα sinβ. 5 derive from the sum and difference identities as follows. cos a + cos b = cos ( a + b 2 + a − b 2 ) + cos ( a + b 2 − a − b 2 ) = ( cos a + b 2 cos a − b 2 − sin a + b 2 sin a − b 2 ) + ( cos a + b 2 cos a − b 2 + sin a + b 2 sin a − b 2 ) =2 cos a + b 2 cos a − b 2 . cos a − cos b = cos ( a + b 2 + a − b 2 ) − cos ( a + b 2 − a − b 2 ) = ( cos a + b 2 cos a − b 2 − sin a + b 2 sin a − b 2 ) − ( cos a + b 2 cos a − b 2 + sin a + b 2 sin a − b 2 ) = − 2 sin a + b 2 sin a − b 2 . sin a + sin b = sin ( a + b 2 + a − b 2 ) + sin ( a + b 2 − a − b 2 ) = ( sin a + b 2 cos a − b 2 + cos a + b 2 sin a − b 2 ) + ( sin a + b 2 cos a − b 2 − cos a + b 2 sin a − b 2 ) =2 sin a + b 2 cos a − b 2 . sin a − sin b = sin ( a + b 2 + a − b 2 ) − sin ( a + b 2 − a − b 2 ) = ( sin a + b 2 cos a − b 2 + cos a + b 2 sin a − b 2 ) − ( sin a + b 2 cos a − b 2 − cos a + b 2 sin a − b 2 ) = 2 cos a + b 2 sin a − b 2 . 6 5. [20] (1) [10] Solve the equation for t on the interval [0, 2π). csc2 t − 2 cot t = 0. Solution: Since 1 + cot2 t = csc2 t, csc2 t − 2 cot t = 0 ⇔ 1 + cot2 t − 2 cot t = 0 ⇔ cot2 t − 2 cot t + 1 = 0 ⇔ (cot t − 1)2 = 0 ⇔ cot t = 1 ⇔ 1 cot t = 1 ⇔ tan t = 1 ⇔ t = π 4 , 5π 4 . (2) [10] Solve the equation. tan−1(3x + 1) = π 4 . Solution: tan−1(3x + 1) = π 4 ⇔ 3x + 1 = tan π 4 ⇔ 3x + 1 = 1 ⇔ 3x = 0 ⇔ x = 0 .