Exam 2 Solution - Plane Trigonometry | MATH 111, Exams of Trigonometry

Download Exam 2 Solution - Plane Trigonometry | MATH 111 and more Exams Trigonometry in PDF only on Docsity! Exam #2 Solutions · Monday November 20, 2006 MATH 111 · Section 7 · Fall 2006 Name The solutions I’ve given here include my explanations and thus are wordier than I would expect your responses to be. Only problem 13 is a writing problem; for the rest, computational answers suffice. Problem 1. Given the function y = 1 3 sin(4x + 1), find the average value and period. Solution: The average value is what’s added to the sin term, which is zero. For the period, solve the inequality 0 ≤ 4x + 1 ≤ 2π −1 ≤ 4x ≤ 2π − 1 − 1 4 ≤ x ≤ 2π 4 − 1 4 . The period is the right-hand side minus the left-hand side, which is π/2. Problem 2. Given the following graph, find the maximum value, minimum value, and amplitude. 0 1 2 3 4 5 6 7 8 9 10 −2 −1 0 1 2 3 4 5 6 Solution: Read off the values max = 5, min = -1, and amplitude = 3. Or, amplitude is (max−min)/2 = (5 + 1)/2 = 3. Problem 3. Write an equation for a sine function which passes through all the points in the following data table: x 0 1 2 3 4 5 y −1 3 7 3 −1 3 Solution: First graph the function: 0 1 2 3 4 5 −2 −1 0 1 2 3 4 5 6 7 8 Now read off the amplitude, average value, period, and phase shift from the graph and write down y = avg. val. + amplitude · sin ( 2π period (x − phase shift) ) y = 3 + 4 sin ( 2π 4 (x − 1) ) = 3 + 4 sin (π 2 (x − 1) ) . 2 Problem 4. Which of the following correctly represents cos(−θ) + sin(−θ)? (A) cos(θ) + sin(θ) (B) cos(θ) − sin(θ) (C) − cos(θ) + sin(θ) (D) − cos(θ) − sin(θ) (E) None of these. Solution: Since cosine is an even function and sine is an odd function, we have, for all θ, cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). So, the answer is B. Problem 5. Verify the following trigonometric identity. (Hint: use a sum identity and the Pythagorean identity.) cos(2x) = 1 − 2 sin2(x). Solution: The sum identity for cosine gives us cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2(x) − sin2(x). You’re asked to get the right-hand side looking like 1− 2 sin2(x), but there is a cos2(x) term in there to be gotten rid of. Fortunately, we know from the Pythagorean identity that cos2(x) + sin2(x) = 1. Using this, we can put cos2(x) = 1 − sin2(x) into the above, picking up from where we left off, to obtain cos(2x) = cos2(x) − sin2(x) = (1 − sin2(x)) − sin2(x) = 1 − 2 sin2(x). Problem 6. Determine the exact value of sin(7π/12) using sum and/or difference identities. Solution: This is perhaps easier in degrees, perhaps not. Using degrees, we have 7π 12 = 7π 180◦ 12π = 7 · 15◦ = 105◦. Now, 105◦ = 60◦ + 45◦ so we can take advantage of known values for sine and cosine at 60◦ and 45◦, along with the sum identity for sine: sin(60◦ + 45◦) = sin(60◦) cos(45◦) + cos(60◦) sin(45◦) = √ 3 2 √ 2 2 + 1 2 √ 2 2 = √ 6 + √ 2 4 . Using radians, we have 7π 12 = 4π 12 + 3π 12 = π 3 + π 4 so we can take advantage of known values for sine and cosine at π/3 and π/4, along with the sum identity for sine: sin(π/3 + π/4) = sin(π/3) cos(π/4) + cos(π/3) sin(π/4) = √ 3 2 √ 2 2 + 1 2 √ 2 2 = √ 6 + √ 2 4 . Remember that switching between radians and degrees changes the units of the input to the trigono- metric functions, but doesn’t change the output. So, we get the same answer either way. 5 Problem 11. Find the exact value of cos(sin−1(−2/3)). Solution: Remember that trig functions take angles to numbers, while their inverse functions take numbers to angles. So, sin−1(−2/3) is an angle. Which angle is it? The sin−1 function has a range of −90◦ to 90◦; −2/3 is negative so we are looking for an angle θ with negative height, where sin(θ) = −2/3. We can draw a cartoon to help us get the math right: - 6 Z Z Z Z Z Z r = 1 y = −2 3 x =? On the unit circle, r = 1 and so cos(θ) = x/r = x/1 = x. We can use the Pythagorean theorem, x2 + y2 = 1, to find x2 + ( −2 3 )2 = 1 x2 + 4 9 = 1 x2 = 1 − 4 9 = 5 9 x = ± √ 5 3 . All angles given back by the sine function are between −90◦ and 90◦ and so are in quadrants I and IV, where x is non-negative. So, we choose the positive square root to get x = √ 5 3 . Problem 12. Find the exact value of arcsin(sin(135◦)). Solution: From chapter 1, sin(135◦) is one of our well-known values, namely, √ 2/2. Now remember that trig functions map angles to numbers, while their inverse functions map numbers to angles. There are, of course, many angles whose sine is √ 2/2, and functions (in order to be functions) can have only one output value. We define the inverse sine function to map numbers from −1 to 1 to angles between −90◦ and 90◦. So, arcsin( √ 2/2) = 45◦. Problem 13. Writing question: Why is it that tan−1(2) is defined, while sin−1(2) and cos−1(2) are undefined? Solution: The domain of the tangent function is (−∞, +∞) while the domain of the sine and cosine functions is [−1, 1]; 2 is in the former but not the latter. The above sentence is an acceptable answer, of acceptable length. For more explanation of why that, in turn, is true, though: • Remember that trig functions map angles to numbers, and their inverse functions map numbers to angles. 6 • Also remember that when we draw an angle in standard position, then pick a point (x, y) which is on the terminal edge of the angle and also on the unit circle, the y coordinate (the height) of that point is the sine of the angle. Likewise, the x coordinate of that point is the cosine of the angle. The tangent of the angle is y/x which is the slope of the hypotenuse. The y and x coordinates are on the unit circle, so they’re confined between −1 and 1. The slope, on the other hand, can get as steep as we want. • Last, remember that when we compute sin−1 of a number, we’re asking: What’s our favorite angle whose sine is that number? Similarly for the cos−1 and tan−1 functions. So, since sine and cosine don’t have value 2 for any angle, we won’t be able to find an angle with sine or cosine equal to 2. On the other hand, since the tangent function takes on all real values, we will be able to find an angle with tangent equal to 2.