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Derivatives of Trigonometric Functions The basic trigonometric limit: sin x x lim = 1 = lim Theorem: x→0 x →0 sin x x (x in radians) Note: In calculus, unless otherwise noted, all angles are measured in radians, and not in degrees. This theorem is sometimes referred to as the small-angle approximation because it really says that, for very small angles x, sin x ≈ x. cos x = 1 . Note: Cosine behaves even better near 0, where lim x →0 ex. Show that lim x →0 cos x − 1 =0 x − sin 2 x cos 2 x − 1 cos x − 1 cos x − 1 cos x + 1 = lim = lim lim = lim ⋅ x →0 x →0 x x cos x + 1 x→0 x(cos x + 1) x→0 x(cos x + 1) = lim x→0 − sin x sin x sin x sin x 0 ⋅ lim = − lim ⋅ lim = − (1) = 0 x →0 cos x + 1 x →0 x x x→0 cos x + 1 1 +1 ex. Evaluate lim x→0 lim x→0 sin 2 x 5x sin 2 x 1 sin 2 x 2 2 sin 2 x = lim ⋅ = lim 5x 5 x →0 x 2 5 x →0 2 x The idea above is to match the angle in the sine function with the denominator. We’ll then apply the basic trigonometric limit. To do so, first we substitute θ = 2x. Note that as x approaches 0, so does θ. Hence, 2 sin 2 x 2 sin θ 2 2 lim = lim = ⋅1 = 5 x →0 2 x 5 θ →0 θ 5 5 ex. Evaluate lim x→0 lim x→0 sin 4 x sin 3 x sin 4 x sin 4 x sin 4 x x sin 4 x x x = lim ⋅ = lim ⋅ = lim ⋅ lim x →0 sin 3 x sin 3 x x→0 sin 3 x x x→0 x sin 3 x x→0 x Repeat the same trick as in the previous example, let θ = 4x and λ = 3x. Both θ and λ approach 0 when x does. Then apply the theorem twice. = lim x →0 = sin 4 x 4 x 3 4 sin 4 x 3x ⋅ ⋅ lim ⋅ = lim ⋅ lim x 4 x→0 sin 3 x 3 3 x→0 4 x x→0 sin 3 x sin θ 4 λ 4 4 lim ⋅ lim = ⋅1⋅1 = λ →0 sin λ 3 θ →0 θ 3 3 In fact, after doing a few examples like those, we can see a (very nice) pattern. To sum it up: Suppose m and n are nonzero real numbers, then lim sin mx m = nx n lim mx m = sin nx n lim sin mx m = sin nx n x→0 x →0 x →0 (Trivially, we also have: lim x →0 mx m = .) nx n ex. Evaluate lim x→0 lim x→0 = tan 7 x 2x tan 7 x 1 sin 7 x 1 sin 7 x 1 1 sin 7 x 1 = lim ⋅ = lim ⋅ = lim ⋅ lim x →0 2 x cos 7 x x →0 cos x 2x 2 x →0 x cos x 2 x→0 x 1 7 1 7 ⋅ ⋅ = 2 1 1 2 Recall that since cos x is continuous everywhere, the direct substitution property applies, therefore, 1 1 1 1 = = = =1 x → 0 cos x lim cos x cos 0 1 lim x →0 Now, the main topic -Derivatives of Trigonometric Functions ex. What is the derivative of sin x? Start with the limit definition of derivative: d sin( x + h) − sin x [sin x cos h + sin h cos x] − sin x = lim sin x = lim h →0 h →0 dx h h = lim h →0 sin x cos h − sin x sin h cos x sin x(cos h − 1) sin h + lim = lim + lim ⋅ cos x h →0 h →0 h →0 h h h h = lim sin x ⋅ lim h →0 Therefore, h →0 cos h − 1 sin h + lim ⋅ lim cos x = sin x ⋅ (0) + (1) cos x = cos x h →0 h h h→0 d sin x = cos x dx ex. Find the derivative of csc x. d d sin x 1 − 1 sin x d 1 d = sin x ⋅ 0 − 1 ⋅ cos x dx dx csc x = = 2 (sin x) sin 2 x dx sin x dx = − cos x 1 cos x =− ⋅ = − csc x cot x 2 sin x sin x sin x Therefore, d csc x = − csc x cot x dx The complete list of derivatives of trigonometric functions: 1. d sin x = cos x dx 2. d cos x = − sin x dx 3. d tan x = sec 2 x dx 4. d sec x = sec x tan x dx 5. d cot x = − csc 2 x dx 6. d csc x = − csc x cot x dx ex. Differentiate f (x) = sec x + 5 csc x f ′(x) = sec x tan x + 5( −csc x cot x) = sec x tan x − 5 csc x cot x ex. Differentiate f (x) = x2 cos x − 2x sin x − 3 cos x f ′(x) = [x2(−sin x) + (2x) cos x] − 2[x (cos x) + (1)sin x] − 3(−sin x) = − x2 sin x + 2x cos x − 2x cos x − 2sin x + 3sin x = − x2 sin x + sin x ex. Differentiate s (t ) = s ′(t ) = sin t 1 − cos t (1 − cos t )(cos t ) − (sin t )(0 − (− sin t )) (1 − cos t ) 2 cos t − cos 2 t − sin 2 t cos t − (cos 2 t + sin 2 t ) = = (1 − cos t ) 2 (1 − cos t ) 2 = cos t − 1 1 − (1 − cos t ) −1 = = = (1 − cos t ) 2 (1 − cos t ) 2 1 − cos t cos t − 1 ex. Simple Harmonic Motion Suppose the oscillating motion (in meters) of a weight attached to a spring is described by the displacement function s(t) = 2 cos t + sin t Find its velocity and acceleration functions, and its speed and acceleration at t = π/2 sec. Velocity: v(t) = s′(t) = −2 sin t + cos t Acceleration: a(t) = v′(t) = −2 cos t − sin t Its speed when t = π/2 is │v(π/2)│ = │−2 sin (π/2) + cos (π/2) │ = │−2 + 0│ = 2 (m/sec) Its acceleration at the same time is a(π/2) = −2 cos (π/2) − sin (π/2) = 0 − 1 = −1 (m/sec2)