Derivatives of Trigonometric Functions

Size: px

Start display at page:

Download "Derivatives of Trigonometric Functions"

Angel Wiggins
5 years ago
Views:

1 Derivatives of Trigonometric Functions In this section, I ll iscuss its an erivatives of trig functions. I ll look at an important it rule first, because I ll use it in computing the erivative of sin. If you graph y sin an y, you see that the graphs become almost inistinguishable near : That is, as, sin. This approimation is often use in applications e.g. analyzing the motion of a simple penulum for small isplacements. I ll use it to erive the formulas for ifferentiating trig functions. In terms of its, this approimation says sin. (Notice that plugging in gives.) A erivation requires the Squeeze Theorem an a little geometry. What I ll give is not really a proof from first principles; you can think of it as an argument which makes the result plausible. y tan sin

2 I ve rawn a sector subtening an angle θ insie a circle of raius. (I m using θ instea of, since θ is more often use for the central angle.) The inner right triangle has altitue sinθ, while the outer right triangle has altitue tanθ. The length of an arc of raius an angle θ is just θ. (I ve rawn the picture as if θ is nonnegative. A similar argument may be given if θ <.) Clearly, sinθ θ tanθ. Divie through by sinθ: As θ, θ sinθ cosθ. just plug in. By the Squeeze Theorem, cosθ θ θ sinθ. Taking reciprocals, I get sinθ. θ θ Eample. Compute sin7. Plugging in gives. I have to o some more work. The it formula has the form sin. In this eample, 7. In orer to apply the formula, I nee 7 on the bottom of the fraction as well as insie the sine: They must match. I can t o much about the 7 insie the sine, but I can make a 7 on the bottom easily using algebra: Let u 7. As, u 7. So sin7 sin sin sinu 7 7. u u I ll often omit writing a substitution like u 7. Once I see that I have something of the form sin where, I know it has it. 5+sin3 Eample. Compute tan4 7cos2. Plugging in gives. The iea here is to create terms of the form sin, to which I can apply my it rule. I ll escribe the steps I ll take first, then o the computation. (a) I ll convert the tangent term to sine an cosine. This is because my funamental rule involves sine, an I also know that cos cos as (so cosine terms aren t much of an issue). 2

3 (b) I ll ivie all the terms on the top an the bottom by. This is in preparation for making terms of the form sin. (c) I ll use the trick I use earlier to fi up numbers so the sine terms all have the form sin, where the thing insie the sine an the thing on the bottom match. Here s the computation: 5+sin3 tan4 7cos2 5+sin3 sin4 cos4 7cos2 5+sin3 sin4 cos4 7cos2 5 + sin3 sin4 cos4 7cos2 sin3 sin4 cos4 7cos sin sin3 3 cos4 7cos As, the terms sin4 an sin3 both go to by the sine it formula. On the other han, the 4 3 terms cos2 an cos4 both go to, since cos an cos is continuous. Eample. (a) Compute cos 2. (b) Compute cos( 6 ) 2. (a) Plugging in gives. The it may or may not eist. The iea is to use a trig ientity (cos) 2 (sin) 2 to change the cosines into sines, so I can use my sine it formula. It is kin of like multiplying the top an bottom of a fraction by the conjugate to simplify a raical epression. cos cos 2 2 +cos +cos (cos) 2 2 (+cos) (sin) 2 2 (+cos) ( ) 2 ( sin +cos ) (b) If you raw the graph near with a graphing calculator or a computer, you are likely to get unusual 3

4 results. Here s the picture: The problem is that when is close to, both 6 an 2 are very close to proucing overflow an unerflow. Actually, the it is easy: Let y 6. When, y, so cos( 6 ) cosy 2 y y 2 2. For the last step, I use the result from the previous problem. Eample. Compute tan7 tan2. If you set, you get. Sigh. I ll see what I can tell from the graph: It looks as thought the it is efine, an the picture suggests that it s aroun 3.5. First, I ll break the tangents own into sines an cosines: tan7 tan2 sin7 cos2 cos7 sin2. 4

5 Net, I ll force the sinθ form to appear. Since I ve got sin7 an sin2, I nee to make a 7 an a 2 θ to match: sin7 cos2 cos7 sin2 7 2 sin7 2 cos2 7 sin2cos7. Now take the it of each piece: sin7 7, 2 sin2, cos2 cos7. The it of a prouct is the prouct of the its: 7 2 sin7 7 2 cos2 sin2cos Derivatives of trig functions. I ll begin with a lemma I ll nee to erive the erivative formulas. cosh Lemma.. h h Proof. Proposition. cosh cosh cosh+ h h h h cosh+ h ( )( ) sinh sinh h h h cosh+ (a) sin cos. (b) cos sin. (c) tan (sec)2. () sec sectan. (e) cot (csc)2. (f) csc csccot. Proof. To prove (a), I ll use the sine it formula I ll also nee the angle aition formula for sine: sinθ. θ θ (cosh) 2 h(cosh+) h (sinh)2 h(cosh+) ( ). + sin(a+b) sinacosb +sinbcosa. 5

6 Let f() sin. Then f f(+h) f() sin(+h) sin sincosh+sinhcos sin () h h h h h h cosh sinh cosh (sin) +(cos) (sin) +cos. h h h h h h The first term goes to by the preceing lemma. Hence, f () cos. That is, sin cos. To erive the formula for cosine, I ll use the angle aition formula for cosine: Let f() cos. Then cos(a+b) cosacosb sinasinb. f f(+h) f() cos(+h) cos coscosh sinsinh cos () h h h h h h (coscosh cos) (sinsinh) (cos)(cosh ) (sinsinh) h h h h (cos)cosh sinh (sin) (cos) (sin) sin. h h h h I won t o the proofs for the remaining trig functions. The iea is to write tan sin cos, cot cos sin, sec cos (cos), csc sin (sin). Then you can use the erivative formulas for sine an cosine together with the quotient rule or the chain rule to compute the erivatives. As an eample, I ll erive the formula for cosecant: csc sin (sin) 2 cos sin cos sin csccot. Eample. Compute the following erivatives. (a) ( 3 3 +cos ). (b) (sin). (c) 4sin+3 5+2cos. () (+sin)(2 tan). (e) 2 sec 3+4csc. 6

7 (a) (b) (c) () (e) ( 3 3 +cos ) 9 2 sin. (sin) ()(cos)+(sin)() cos+sin. 4sin+3 5+2cos (5+2cos)(4cos+3) (4sin+3)( 2sin) (5+2cos) 2. (+sin)(2 tan) (+sin)(2 (sec) 2 )+( 2 tan)( cos). 2 sec 3+4csc (3+4csc)( sectan) (2 sec)( 4csccot) (3+4csc) 2. Eample. For what values of oes f() +sin have a horizontal tangent? f () +cos. So f () where cos. In the range 2π, this happens at π. So f () for π +2nπ, where n is any integer. c 28 by Bruce Ikenaga 7

Integration by Parts

Integration by Parts 6-3-207 If u an v are functions of, the Prouct Rule says that (uv) = uv +vu Integrate both sies: (uv) = uv = uv + u v + uv = uv vu, vu v u, I ve written u an v as shorthan for u an