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1 Section 2.6 Derivative of Trigonometric Functions Some Important Trigonometric Limits To be able to calculate the derivative of the trig functions we need to be able to evaluate : sin(θ + h) − sin(θ) h→0 h (sin(θ))0 = lim cos(θ + h) − cos(θ) h→0 h First let’s recall couple of facts about the limits of trig functions from Chapter 1. (cos(θ))0 = lim 1) lim sin(θ) = 0 and θ→0 2) lim cos(θ) = 1 θ→0 These facts follows from the fact that sine and cosine are continuous functions. We will start evaluating the limit for the derivative of the sine function at θ=0 sin(h) 0 sin(0 + h) − sin(0) = lim = h→0 h h 0 So this limit is indeterminate form. In your exam 1 you have been told that this limit is equal to 1. Now using geometry we will first prove this fact. (sin(0))0 = lim h→0 2 Lemma lim θ→0 sin(θ) θ =1 Proof: Assume that the angle 0 < θ < π2 . Figure above shows a sector of a circle with center O=(0,0), central angle θ, and radius 1. The line PK is drawn perpendicular to OR. By definition of radian measure, we have arc P R = θ. Also, |P K| = |OP | sin(θ) = sin(θ) so P = (cos(θ), sin(θ)). Also note that | {z } =1 the triangles 4OP K and 4OQR are similar. So tan(θ) = QR QR sin(θ) = = = QR cos(θ) OR 1 So Q = (1, tan(θ)). Now we will compare the areas. It is clear from the figure that the area of the triangle OPR is less than the area of the sector OPR of the unit circle which is less than the area of the big triangle OQR area of 4 OQR > area of the sector OP R > area of 4 OP R 1 (1) tan(θ) > 2 π(1)2 | {z } θ 2π |{z} · area of unit circle the fraction of the circle in this sector So, tan(θ) θ sin(θ) > > 2 2 2 Divide each side of the inequality by 1 2 sin(θ) 1 > (1) sin(θ) 2 3 1 θ > >1 cos(θ) sin(θ) Take reciprocals: sin(θ) <1 θ Now since lim cos(θ) = 1 and lim 1 = 1 by Squeeze Theorem cos(θ) < θ→0 θ→0 sin(θ) =1 θ→0 θ Now if you go back to the derivative of sin(x) at x = 0; lim d(sin(x)) sin(h) |x=0 = sin0 (0) = lim =1 h→0 dx h Lemma lim θ→0 cos(θ)−1 θ =0 Proof Note first that the first attempt to plug in θ = 0 gives you the indeterminate form 00 . To evaluate this limit we will use the result we got in the above lemma as well as the trig property cos2 (θ) + sin2 (θ) = 1 ⇒ sin2 (θ) = cos2 (θ) − 1. cos(θ) − 1 cos(θ) + 1 cos(θ) − 1 = lim · θ→0 θ→0 θ θ cos(θ) + 1 2 cos (θ) − 1 = lim θ→0 θ(cos(θ) + 1) sin2 (θ) = lim θ→0 θ(cos(θ) + 1) sin(θ) sin(θ) = lim ( · ) θ→0 θ cos(θ) + 1 sin(θ) sin(θ) = lim · lim θ→0 θ→0 cos(θ) + 1 θ =1·0=0 lim Now let’s first try to find the derivative of cos(x) at x = 0; cos0 (0) = lim h→0 cos(h) − 1 cos(0 + h) − cos(0) = lim =0 h→0 h h 4 Trigonometric Derivatives Theorem d [sin(x)] = cos(x) dx Proof sin(x + h) − sin(x) h→0 h sin(x) cos(h) + sin(h) cos(x) − sin(x) = lim h→0 h µ ¶ sin(x) cos(h) − sin(x) sin(h) cos(x) = lim + h→0 h h µ ¶ µ ¶ cos(h) − 1 sin(h) = sin(x) lim + cos(x) lim h→0 h→0 h h = sin(x) · 0 + cos(x) · 1 = cos(x) (sin(x))0 = lim Theorem d [cos(x)] = − sin(x) dx Proof Similar like the above one. This is a good exercise for you to try. Once you have the derivative of sin(x) and cos(x), we have all the other trig derivatives: Theorem d [tan(x)] = sec2 (x) dx Proof · ¸ d sin(x) d [tan(x)] = (Use Quotient Rule) dx dx cos(x) cos x cos x − sin x(− sin x) = [cos2 x] 2 cos x + sin2 x = cos2 x 1 = = sec2 x cos2 x 5 You can similarly show the following is true: d [cot(x)] = − csc2 (x) dx d [sec(x)] = sec(x) tan(x) dx d [csc(x)] = − csc(x) cot(x) dx Example Find y 0 for y = tan(x)(x3 − ex + 1) y 0 = sec2 (x)(x3 − ex + 1) + tan(x)(3x2 − e) Example Find y 0 for y = sin(cos(tan x)). d [cos(tan x)] dx d = cos(cos(tan x))[− sin(tan x) [tan x]] dx y 0 = cos(cos(tan x)) · = cos(cos(tan x)) · (− sin(tan x)) · (sec2 x) Example Find the points on the curve y = horizontal. cos(x) 2+sin(x) at which the tangent is (2 + sin x)(− sin x) − (cos x)(cos x) (2 + sin x)2 −2 sin x − sin2 x − cos2 x = (2 + sin x)2 −2 sin x − 1 = (2 + sin x)2 y0 = Tangent line is horizontal when y 0 = 0 y0 = −2 sin x − 1 1 7π 11π = 0 ⇒ −2 sin x−1 = 0 ⇒ sin x = − ⇒ x = +2πk or x = +2πk 2 (2 + sin x) 2 6 6