AM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h

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1 AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x + γ) (.) where γ = h. We re going to show (.). he proof of (.) is left s n homework. Recll the sum-to-product identity ( ) ( ) θ ϕ θ + ϕ sin θ sin ϕ = sin cos If we use it with θ = x + h nd ϕ = x, we obtin sin(x + h) sin x = ( ) ( ) h x + h h h sin cos (.) (.4) which is just equivlent to (.). Exercise.. Prove or disprove ech of the following sttements.. For ll x, we hve sin x sin x.. For every x, there is y such tht cos(x + y) = cos x + cos y.. here is n x such tht sin(x + y) = sin x + sin y for ll y. 4. here is y such tht y sin x dx = sin y. his sttement is clerly flse. ke indeed x = π, then sin 4π = = sin π. -

2 -. his one is flse s well. ke x =, then you hve cos(x + y) = cos(y) + cos(y) = cos(x) + cos(y) (.5). his sttement is true, nd we prove it exhibiting the right x, tht is. With this choice, sin(x + y) = sin y = + sin y = sin x + sin y (.6) note tht (.6) is vlid for ny y. 4. his lst sttement is true. We prove it clculting the integrl, y sin x dx = cos(y) (.7) hen we wnt to exhibit y such tht sin y = cos y (.8) his equlity clerly holds for y = π, since sin π = nd cos π eqution (.8) holds for ny y = π + kπ, k Z. =. As mtter of fct, Exercise.. Clculte the integrl b sin x dx for ech if the following vlues of nd b. In ech cse interpret your result in terms of res.. =, b = π 6 nd =, b = π 4. =, b = π nd =, b = π. =, b = π nd =, b = π 4. =, b = nd = π 6, b = π 4 We will solve only cse no. since the others re completely nlogous nd left s n esy homework. π 6 ( π ) sin x dx = cos 6 = (.9) (.) π 4 ( π ) sin x dx = cos 4 = (.) (.)

3 - Exercise.4. Evlute the following integrls. π (x + sin x) dx. π (x + cos x) dx. π (sin x cos x) dx 4. π 5. π 6. π 7. π sin x cos x dx ( + cos t) dt + cos t dt + cos t dt, with x π 8. x (t + sin t) dt 9. π sin x dx. π cos x dx. π π π (x + sin x) = + cos π = +. π (x + cos x) = π 4 +. π (sin x cos x) = = 4. π sin x cos x = π 4 5. π 6. π ( + cos t) = π + cos t π = (cos x sin x) + π (sin x cos x) = ( ) π 4 ( + cos t) + π ( π cos t) = 7π + 7. We hve to distinguish two cses: () x π nd (b) π < x π. () π (b) π + cos t = π ( π cos t) + ( π + cos t) = 7π + + x + sin x + cos t = π ( π cos t) + π ( + cos t) + ( π cos t) π the computtion of (b) is left s n esy homework. 8. x (t + sin t) = (x ) (x) + cos(x) cos(x ) = x6 x + cos(x) cos(x )

4 -4 9. π sin x = π sin x = here you hve to tke ϕ = nd note tht π sin x = ( ) πϕ ϕ sin x ϕ which in turn is equl to ϕ π sin x by formul (.6) from lst notes.. π cos x = π 6 cos x = gin by mens of (.6). Exercise.5. Use the identity sin t = sin t 4 sin t to deduce the integrtion formul sin t dt = ( + sin x) cos x By linerity with respect to the integrnd we hve which is our gol. sin t = 4 [ sin t x sin t = [ sin t sin t 4 = [ 8 x sin t x sin t 4 x = [ 8 4 ( cos x) (cos x cos x) = [ 8 4 ( cos x) sin x sin x = [ 8 4 ( cos x) 4 sin x cos x = [ cos x( + sin x) Exercise.6. Derive the identity cos t = 4 cos t cos t nd use it to prove tht cos t dt = ( + cos x) sin x

5 -5 We begin verifying the required identity. By mens of ddition formuls cos t = cos t cos t sin t sin t = cos t(cos t sin t) sin t cos t = cos t sin t cos t = cos t ( cos t) cos t = 4 cos t cos t Now we cn use this formul together with the one for sin t from previous exercise. hus cos t = or equivlently, by linerity of the integrl which is our gol. (4 cos t cos t) (.) cos t = (cos t + cos t) 4 = [ x cos t + sin x 4 = [ sin x + sin x 4 = [ 4 sin x 4 4 sin x ( = sin x ) sin x ( = + ) cos x sin x (.) Exercise.7. Prove tht if function f is periodic with period > nd integrble on [,, prove tht f(x) dx = f(x) dx R (.4) If = then (.4) is trivil. Consider for simplicity >, the cse < is nlogous. Recll eqution (.6) from previous lecture f(x) dx = f(x) dx + f(x) dx f(x) dx = = f(x) dx + f(x) dx + f(x ) dx f(x) dx f(x) dx f(x) dx

6 -6 Exercise.8. Prove the following integrtion formuls, vlid for : Consider cos(α + t) nd use ngle ddition formul cos(α + t) dt = [sin(α + x) sin α (.5) sin(α + t) dt = [cos(α + x) cos α (.6) cos(α + t) = (cos α cos t sin α sin t) = x cos α cos t x sin α sin t = [cos α sin x sin α( cos x) = [cos α sin x + sin α cos x sin α = [sin(α + x) sin α he proof of (.6) is nlogous nd left s n esy homework. Exercise.9. Prove tht for ll integers n π π Recll formul (.6) from previous lecture. π sin nx dx = n sin nx dx = (.7) cos nx dx = (.8) nπ sin x dx = ( cos(nπ)) n = he proof of (.8) is nlogous nd left s n esy homework.

7 -7 Exercise.. Using (.7)-(.8) nd the ddition formuls for sine nd cosine estblish the following identities, vlid for ll integers m nd n such tht m n nd n, m. π π π Recll the ngle ddition formuls for sine sin nx cos mx dx = (.9) sin nx sin mx dx = (.) cos nx cos mx dx = (.) π π sin nx dx = π (.) cos nx dx = π (.) sin(α + ) = cos α sin + sin α cos sin(α ) = cos α sin + sin α cos from which we obtin, subtrcting the second equlity from the first, herefore, by mens of (.4) nd (.7) we hve π sin nx cos mx dx = sin(α + ) sin(α ) = sin cos α (.4) = [ π Recll now the ngle ddition formuls for cosine from which we esily obtin sin[x(n + m) dx π cos(α + ) = cos α cos sin α sin cos(α ) = cos α cos + sin α sin herefore, by mens of (.5) nd (.8) we hve π sin nx sin mx dx = sin[x(n m) dx cos(α + ) cos(α ) = sin α sin (.5) = [ π cos[(n m)x dx he proof of (.) is nlogous nd left s n esy homework. π cos[(n + m)x dx

8 -8 Eventully, let us prove (.). From (.5) we get π sin nx dx = = = = π π [ π π sin nx sin nx dx π cos[(n n)x dx cos[(n + n)x dx dx he proof of (.) is nlogous nd left s n esy homework.