21.2 Differentiating sin x and cos x 691 Proof that lim h→0 cos(h)−1 h = 0 lim h→0 cos(h) − 1 h = lim h→0 cos(h) − 1 h · (cos(h) + 1) (cos(h) + 1) = lim h→0 cos 2 (h) − 1 h · 1 (cos(h) + 1) = lim h→0 1 − cos 2 (h) h · −1 (cos(h) + 1) = lim h→0 sin 2 (h) h · −1 (cos(h) + 1) = lim h→0 sin(h) h · − sin(h) (cos(h) + 1) But lim h→0 sin(h) h = 1 and lim h→0 − sin(h) (cos(h) + 1) = 0 1 + 1 = 0, so lim h→0 cos(h) − 1 h = (1)(0) = 0. We have now shown that d dx sin x = cos x, as conjectured! The Chain Rule tells us that d dx sin[g(x)] = cos[g(x)]g  (x). Now that we have proven d dx sin x = cos x, the derivative of cos x is easy to tackle by using the fact that cos x and sin x are related to one another by a horizontal shift. Looking back at the graphs of cos x and its derivative it is easy to speculate that the derivative of cos x is − sin x. Proof that the Derivative of cos x is − sin x Observe that sin  x + π 2  = cos x; replacing x by  x + π 2  shifts the sine graph left π 2 units. Similarly, cos  x + π 2  =−sin x; replacing x by  x + π 2  shifts the cosine graph left π 2 units. sin x = f(x) cos x = g(x) x 1 –1 π 2 –π 2 3π 2 –3π 2 π–π 2π–2π x 1 –1 π 2 –π 2 3π 2 –3π 2 π–π 2π–2π Figure 21.8 692 CHAPTER 21 Differentiation of Trigonometric Functions d dx cos x = d dx sin  x + π 2  = cos  x + π 2  · 1 (by the Chain Rule) =−sin x Combining this result with the Chain Rule gives us d dx sin[g(x)] = cos[g(x)] · g  (x) or informally d dx sin[mess] = cos[mess] · [mess]  d dx cos[g(x)] =−sin[g(x)] · g  (x) d dx cos[mess] =−sin[mess] · [mess]  , where mess is a function of x. Using the Chain Rule and either the Product or Quotient Rule allows us to find the derivatives of tan x, csc x, sec x, and cot x. EXERCISE 21.2 Show that d dx tan x = sec 2 x. EXERCISE 21.3 Show that d dx sec x = sec x tan x. ◆ EXAMPLE 21.1 Differentiate the following. (a) y = 3x sin(x 2 ) (b) y = 7 cos 2 (3x + 5) (c) y =  tan(x 2 ) SOLUTIONS (a) This is the product of 3x and sin[mess]. y  = 3 sin(x 2 ) + 3x cos(x 2 )(2x) = 3 sin(x 2 ) + 6x 2 cos(x 2 ) (b) 7 cos 2 (3x + 5) = 7[cos(3x + 5)] 2 , so basically this is 7[mess] 2 and its derivative is 14[mess] · [mess]  , where the mess is cos(3x + 5). Then the Chain Rule must be applied to cos(3x + 5). y  = 14 cos(3x + 5)(− sin(3x + 5)) · (3) =−42 cos(3x + 5) sin(3x + 5) (c) y =  tan(x 2 ) =  tan(x 2 )  1/2 . This is basically [mess] 1/2 , so its derivative is 1 2 [mess] −1/2 · [mess]  . We know mess = tan(stuff ),so[mess]  = sec 2 (stuff ) · (stuff )  . y  = (1/2)  tan(x 2 )  −1/2 sec 2 (x 2 )2x y  = x sec 2 (x 2 )  tan(x 2 ) ◆ 21.2 Differentiating sin x and cos x 693 An Excursion: Polynomial Approximations of Trigonometric Functions We spent a fair amount of energy proving that lim x→0 sin x x = 1. Now that this fact is ours, we can get some mileage out of it. From lim x→0 sin x x = 1 it follows that sin x ≈ x for very small values of x. 9 This approximation is excellent for very small values of x,but as x gets increasingly far from zero the approximation becomes poor and eventually is useless. For example, sin 0.1 = 0.0998334 ≈0.1 and sin 0.02 = 0.0199986 ≈0.02, but sin 3 = 0.1411200 ,which is not close to 3. y y = x x sin x Figure 21.9 Historically the approximation sin x ≈ x for very small values of x was used in ancient times, well before the development of calculus. Using it, sin 1 ◦ can be approximated with a great degree of accuracy. From there, using addition formulas and knowledge about specific triangles, trigonometric tables can be built up. The approximation sin x ≈ x, when used to estimate sin 1 ◦ ,gives sin 1 ◦ = sin  π 180  ≈ π 180 = 0.01745329 , where the actual value of sin 1 ◦ ≈ 0.0174524 The approximation sin x ≈ x, the tangent line approximation of sin x at x = 0, gives an estimate for sin x that is too large if x is positive and too small if x is negative. (See Figure 21.9). This approximation can be improved upon. The approximation sin x ≈ x − x 3 /6, the best cubic approximation to sin x around x = 0, was used well before the seventeenth century. This gives a higher degree of accuracy for x near zero. Using it to approximate sin 1 ◦ gives sin 1 ◦ = sin(π/180) ≈ π/180 − 1 6 (π/180) 3 ≈ 0.0174524064. This is identical to the numerical approximation supplied by a calculator. The third degree polynomial approximation of sin x around x = 0 can be improved upon by using a fifth degree approximation, which can in turn be improved upon by using a seventh degree polynomial. (We use only polynomials of odd degree because sine is an odd function.) As the degree of the polynomial used to approximate sine increases, the accuracy of the approximation near x = 0 increases and the interval around x = 0 for which the approximation is reasonable enlarges as well. Amazingly enough, if we continue along in this way we can come up with an infinite “polynomial” that is exactly equal to sin x everywhere. These polynomial expansions are known as Taylor series, after the British mathematician Brook Taylor. (We take up Taylor series in Chapter 30). 9 This statement is true only if x is given in radians. 694 CHAPTER 21 Differentiation of Trigonometric Functions By the seventeenth century mathematicians were using infinite polynomial expansions of functions, trigonometric and others. The polynomial expansion of sin x is given by sin x = x − x 3 3 · 2 + x 5 5 · 4 · 3 · 2 − x 7 7 · 6 · 5 · 4 · 3 · 2 +···. If we use factorial notation, letting n! = n(n − 1)(n − 2) 3 · 2 · 1, this can be written as sin x = x − x 3 3! + x 5 5! − x 7 7! +···. The first several terms can be used to approximate sin x for x small. For instance, sin(0.1) = 0.1 − (0.1) 3 3! + (0.1) 5 5! −···, so sin(0.1) ≈ 0.1 − (0.1) 3 3! = 0.0998 3 Compare this with sin(0.1) ≈ 0.0998334166 Using one more term of the polyno- mial gives sin(0.1) ≈ 0.1 − (0.1) 3 3! + (0.1) 5 5! ≈ 0.0998334167. A good match! PROBLEMS FOR SECTION 21.2 1. Using the derivatives of sine and cosine and either the Product Rule or the Quotient Rule, show that d dx tan x = sec 2 x. 2. Show that d dx sec x = sec x tan x. 3. Find the first and second derivatives of the following. (a) f(x)=5cos x (b) g(x) =−3sin(2x) (c) h(x) = 0.5 tan x (d) j(x)=2sin x cos x 4. Differentiate the following. (a) y = cos 2 x (b) y = cos(x 2 ) (c) y = x tan 2 x (d) y = sin 3 (x 4 ) (e) y = 7[cos(5x)+ 3] x 5. Consider the function f(x)=e −0.3x sin x. (a) For what values of x does f(x)have its local maxima and local minima? (b) Is f(x)aperiodic function? (c) Sketch the graph of f(x)=e −0.3x sin x. 21.3 Applications 695 (d) What is the maximum value of for e −0.3x sin x for x ≥ 0? At what x-value is this maximum attained? Your answers must be exact, not numerical approximations from a calculator. Give justification that this value is indeed the maximum. Evaluate the following derivatives. u(x) is a differentiable function. 6. (a) d dx sin ( u(x) ) (b) d dx cos(u(x)) (c) d dx u(x)(sin x) 7. (a) d dx u(x)(cos x) (b) d dx tan(u(x)) (c) d dx u(x)(tan x) Evaluate. 8. d dx sin(x 3 + ln 3x) 9. d dx cos 2 (sin x) 10. d dx  1 sin 3 (cos 2x)  11. d dx  sin(2x 3 ) 12. d dx 4 √ 2−cos(x/7) 13. d dx  e 3x cos 2 (7x)  14. Find dy/dx in terms of x and y. sin(xy) + y = y cos x 15. Find y  . (a) y = x sin x (b) y = 3 tan 3 (x 2 ) (c) y = tan  x 3  sec(3x) 16. Why have we been telling you that radians are more appropriate than degrees when using calculus? Suppose x is measured in degrees. Then cos x ◦ = cos  x ◦ πradians 180 ◦  = cos  πx 180  where the argument is now in radians. Find the derivative. Is the derivative − sin x ◦ ? 21.3 APPLICATIONS Optimization ◆ EXAMPLE 21.2 What angle of launch will propel an object (such as a cannonball or a baseball) farthest horizontally? This question is vitally important to engineers and sportsmen alike. If we consider only the force of gravity (ignoring air resistance, the Coriolis effect, etc.), then it can be shown that the path the object will take is a parabola. In Section 20.7 we showed that if an object is 696 CHAPTER 21 Differentiation of Trigonometric Functions launched at ground level at an angle θ and with an initial velocity of v 0 , then the horizontal distance it will travel is given by R(θ) = 2v 2 0 cos θ sin θ g , where g is the acceleration due to gravity. We want to find θ such that R(θ) is maximum. SOLUTION Using the trigonometric identity sin 2θ = 2 sin θ cos θ, R(θ) can be rewritten as R(θ) = v 2 0 sin(2θ) g or v 2 0 g sin(2θ), where g and v 0 are constants. One approach to this problem is to find the critical points of R(θ). For our purposes θ must be between 0 and π/2. Therefore, the critical points of R are the endpoints θ = 0 and θ = π/2, both resulting in R(θ) = 0, and the values of θ between 0 and π/2 such that dR dθ = 0. dR dθ = v 2 0 g (2 cos(2θ)) Setting dR dθ equal to 0 gives 0 = 2 v 2 0 g cos(2θ) 2 v 2 0 g is a constant. 0 = cos(2θ) Let u = 2θ. When 0 ≤ θ ≤ π 2 ,2·0≤2θ≤2· π 2 ,so 0≤u≤π. cos u = 0 when u = π/2. This is the only value of u ∈ [0, π] that satisfies the equation. 2θ = π/2 (Substitute 2θ = u.) θ = π/4 θ = π/4 is a candidate for the maximum. We can show it is actually the maximum by looking at R  (θ),or d 2 R dθ 2 . R  (θ) = 2v 2 0 g d dθ (cos 2θ) = 2v 2 0 g (−2 sin 2θ) = −4v 2 0 g sin 2θ R  (π/4) = −4v 2 0 g sin(π/2) = −4v 2 0 g 21.3 Applications 697 R  (π/4) = −4v 2 0 g < 0, so the graph of R(θ) is concave down at θ = π/4 and R(θ) has a maximum at π 4 . Therefore, an angle of π 4 radians, or 45 ◦ , is the angle that will give the greatest horizontal distance. NOTE An alternative approach would be to do this optimization entirely without calculus. R(θ) = v 2 0 sin(2θ) g ; we need to find the angle θ ∈  0, π 2  to make sin(2θ) greatest. Either this can be calculated directly, or we can again let u = 2θ. If θ ∈  0, π 2  , then u = 2θ ∈ [0, π]. On [0, π] sin u is maximum at u = π 2 ,soθ= u 2 = π 4 . We have shown that for any fixed initial speed the projectile will travel the farthest horizontal distance if it is launched at a 45 ◦ angle. ◆ ◆ EXAMPLE 21.3 A lighthouse is located 3 kilometers away from a long, straight beach wall. The beacon of light is rotating steadily at a rate of 1 1 2 revolutions per minute. (a) A lone soul is sitting on the beach wall 5 kilometers from the lighthouse, staring into the sea and contemplating the universe. At what rate is the ray of light moving along the beach wall when it passes the thinker? (b) At what point along the beach wall is the beam moving most slowly? SOLUTION (a) Begin with a picture. Do this on your own, thinking carefully about what is known and what you are trying to find. 10 Then compare your work with what is given below. beach wall lighthouse 3 x z θ lone soul lighthouse snapshot 3 4 5 θ Figure 21.10 What We Want: dx dt , the rate at which the ray of light is moving along the beach wall when z = 5. What We Know: The light makes 1 1 2 revolutions per minute, so it goes through 3 2 revolutions minute · 2π radians 1 revolution = 3π radians minute . dθ dt = 3π radians/minute Strategy We know dθ dt and we want to find dx dt . Our strategy is to write an equation relating θ and x. 11 We then differentiate with respect to time to get a relationship between dθ dt and 10 A common error is to assume the beam of light has a fixed length. 11 x varies. It is not always 4. It is only after we have differentiated that we can substitute in values for quantities that vary with time. 698 CHAPTER 21 Differentiation of Trigonometric Functions dx dt . We try to find a trigonometric function involving only sides we know or are concerned about. x 3 = tan θ d dt  x 3  = d dt (tan θ) Differentiate each side with respect to time. 1 3 dx dt = (sec 2 θ) dθ dt x and θ are functions of t, so use the Chain Rule. dx dt = 3(sec 2 θ) dθ dt Solve for dx dt , the rate we want to find. Now we can substitute the values we know. We know dθ dt = 3π.Tofind sec θ, we use the fact that z = 5 at the moment in question. sec θ = hyp adj = 5 3 Therefore, at the moment when the beam passes the thinker, z = 5 and dx dt = 3(sec 2 θ) dθ dt = 3  25 9  · 3π = 25π miles per minute. lone soul lighthouse Figure 21.11 (b) At what point along the cliffs is the spot of light moving most slowly? For our purposes, the angle θ must be between − π 2 and π 2 or the light will not be shining on the cliffs. We know the rate at which the spot of light moves: dx dt = 3(sec 2 θ) dθ dt = 3 · 3π(sec 2 θ) radians/minute. To find the place where the light moves most slowly, we want to find the angle θ that minimizes dx dt . Since sec 2 θ = 1 cos 2 θ , we want to make cos 2 θ as large as possible. This will occur when cos θ is at its maximum, which occurs at θ = 0, or when the light is directly across from the lighthouse, when the beam is shortest. Notice that the further away an object on the beach wall is from the lighthouse, the faster the beam will sweep past it. ◆ 21.3 Applications 699 PROBLEMS FOR SECTION 21.3 1. Let f(x)=x+2sin x. (a) Find all of the critical points. (b) Where is f(x)increasing? Decreasing? (c) Where does f(x)have local maxima? Local minima? (d) Does f(x) have global maxima? Global minima? If so, what are the absolute maximum and minimum values? (e) Where is f(x)concave up? Concave down? (f) Sketch a graph of f(x). 2. Consider the function f(x)=−cos x + 1 2 sin 2x. (a) Explain how you can tell that f is periodic with period 2π. (b) Find and classify all the critical points of f on the interval [0, 2π]. Do the trigonometric “algebra” on your own, then check your answers using a graphing calculator. (Hint: You ’ ll get a cos 2x that you’ll need to rewrite.) For Problems 3 through 6, graph f on the interval [0, 2π] labeling the x-coordinates of all local extrema. 3. f(x)=cos x + √ 3 sin x 4. f(x)=cos x − sin x 5. f(x)=cos 2x − 2 cos x 6. f(x)=e x sin x 7. Use a tangent line approximation to approximate the following. In each case, use concavity to determine whether the approximation is larger or smaller than the actual value. Then compare your results with the approximations given by a calculator or computer. (a) sin 0.2 (b) sin 0.1 (c) sin 0.01 (d) sin(−0.1) 8. Creme Fraiche and Caveat are battling their way to the finish line in the last leg of horseracing’s Triple Crown. At the finish line, 30 feet away from the track itself, is a camera that is focused on the leading horse who is moving down the stretch at a rate of 46 feet per second. At what rate is the camera rotating when the lead horse is 50 feet from the finish line? 9. Verify that sec x has local minima at x =2πk and local maxima at x = π + 2πk (k an integer) by identifying its critical points and using the second derivative test for maxima and minima. 10. Verify that tan x has points of inflection at x = πk, k an integer, by showing that the sign of its second derivative changes at these points. 700 CHAPTER 21 Differentiation of Trigonometric Functions 11. Let f(x)=3cos x + 2 sin x. (a) What is the period of f ? (b) What are the maximum and minimum values of f ? 12. If we ignore air resistance, a baseball thrown from shoulder level at an angle of θ radians with the ground and at an initial velocity of v 0 meters per second will be at shoulder level again when it is v 2 0 sin(2θ) g meters away. g is the acceleration due to gravity (9.8 m/sec 2 ). (a) Express the maximum distance the baseball can travel (from shoulder level to shoulder level) in terms of the initial velocity. (b) The fastest baseball pitchers can throw about 100 miles per hour. How far would such a ball travel if thrown at the optimal angle? (Note: 1 mile = 5280 feet and 1 meter ≈ 3.28 feet.)(*) 13. A policewoman is standing 80 feet away from a long, straight fence when she notices someone running along it. She points her flashlight at him and keeps it on him as he runs. When the distance between her and the runner is 100 feet he is running at 9 feet per second. At this moment, at what rate is she turning the flashlight to keep him illuminated? Include units in your answer. 14. A sewage gutter is to be constructed from a piece of sheet metal 8 feet long and 4 feet wide by folding up a 1-foot strip on each side. Denote by θ the angle between the sides and the vertical, as shown in the figure below. What angle θ will result in a sewage gutter of maximum volume? 1 foot 1 foot θ θ 15. A lookout tower is located 0.5 kilometers from a line of warehouses. A searchlight on the tower is rotating at a rate of 6 revolutions per minute. How fast is the beam of light moving along the wall of warehouses when it passes by a window located 1 kilometer from the tower? 16. Graph f(x)=2 cos x . (a) Is the function periodic? If so, what is its period? (b) What is its maximum value? Its minimum value? Give exact answers. 17. Let f(x)=−cos x and g(x) = sin x. (a) What is the maximum distance between these two curves on the interval [− π 4 , 3π 4 ]? (b) What is the point of intersection of the tangent lines to these curves at the points from part (a) where the curves are farthest apart? Does this answer surprise you? Explain. . are the maximum and minimum values of f ? 12. If we ignore air resistance, a baseball thrown from shoulder level at an angle of θ radians with the ground and at an initial velocity of v 0 meters. 3π radians minute . dθ dt = 3π radians/minute Strategy We know dθ dt and we want to find dx dt . Our strategy is to write an equation relating θ and x. 11 We then differentiate with respect to time to get a relationship. graph of R(θ) is concave down at θ = π/4 and R(θ) has a maximum at π 4 . Therefore, an angle of π 4 radians, or 45 ◦ , is the angle that will give the greatest horizontal distance. NOTE An alternative