Cos3x 2cos2x 1 2sinx sin2x
[PDF File]Answers to Maths B (EE1.MAB) exam papers
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2. (i) 1 12, (ii) −1, (iii) 9 2. 3. (i) (a) 1 2 (x 2−1/x ), (b) 0 (ii) 2sinh−1(1) or 2ln(1+ √ 2) 4. (i) y = Aex4/4, (ii) y = 2 1+e−2x, (iii) y = 1 2 + c (x+1)2. Depending on how you do the calculations you might instead get y = x2/2+x (x+1)2 + d (x+1)2 which is equivalent. 5. (i) y = e−x(−2cos2x− 1 2 sin2x), (ii) y = Ae−2 x+Be ...
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1 cos2 E tan sin sin f) 1 cosa sina F 1 cosa sina Giải: a) sin3xcos5x sin3xcos5x sin(3x 5x) sin( 2x) sin2x A cosx cosx cosx cosx 2sinxcosx 2sinx cosx b) sinx sin4x sin7x (sin7x sinx) sin4x 2sin4xcos3x sin4x B cosx cos4x cos7x (cos7x cosx) cos4x 2cos4xcos3x cos4x = sin4x(2cos3x 1) sin4x tan4x cos4x(2cos3x 1) cos4x c) 2sin2 sin4 2sin2 2sin2 ...
PHƯƠNG TRÌNH LƯỢNG GIÁC
1. sin2x+sin6x sin8x =0 2. sinx+sin3x =cosx+cos3x 3. (1+tanx)(1 sin2x)=1 tanx 4. 1+sinx+2cosx+sin2x =0 5. 2cos2x+sin2x+5=8cosx+sinx 6. tan2x = 1+cosx 1 sinx 7. sin 2x cos 2x =sin 23x cos 4x 8. cos5x+sin5x 2sin3x+sinx cosx =0 9. 3cos2x+20cos3xcos2x 10cos5x 1 p sinx =0 10. cos 6x+sin x = 1 8 (5+6cos7xcos3x) 11. sin5x =cosxtan3x 12. 1 sin2x 2sinx ...
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2sinx — 2 sin x 2 sin x cosx sin 2 sinxcosx 2sinxco_sx cosx (2cos2x — 1 Isinx 2 sinxcosx 2 sin xcosx 2 sin x cosx — 2cosx — —secx — 2cosx — 2 sin A 2sinAcos: A — sin.A ... cos3x — cos2r 4 cosx sin3x — sin2x sinx COS 31 — cos2x 2cos2tcosx — sin + sinx — sin 2K 2sin2.scosx cos2x(2cosx — Ij
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= cos2x + sin2x + 2 sin s cosx - 1 = 1 +2 sin x. cosx-l 2 sin x cos x cos2 - cos2x . sin2x = cos2x (1 - sin2x) — cos2x . cos2x C(x) = cos4x (2cos x + sin + (cos x -2sin D(x) = = 4cos2x + sin2x + x . sin X + cos2x + 4sin2 x - x cos x = 5cos2x + 5 sin2x — 5(cos2x + sin2x) E(x) = cos5x + cos3x . sin2x = cos3 (cos2x + sin2x) — cos3x . 1 F(x ...
[PDF File]10 Fourier Series - UCL
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1 2 a 0 + a 1 cosx+a 2 cos2x+a 3 cos3x+... + b 1 sinx+b 2 sin2x+b 3 sin3x+... (1) Multiply (1) by cos(nx),n ≥ 1 and integrate from −π to π and assume it is permissible to integrate the series term by term. Z π −π f(x)cos(nx)dx = 1 2 a 0 Z π −π cos(nx)+a 1 Z π −π cosxcos(nx)dx+a 2 Z π −π cos2xcos(nx)dx+... +b 1 …
[PDF File]Techniques of Integration - Whitman College
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204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate Z sin6 xdx. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. Now we have four integrals to evaluate: Z 1dx = x and Z
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∵ sin5x.cos3x = sin6x.cos2x ⇒ 2sin5x.cos3x = 2sin6x.cos2x ⇒ sin8x + sin2x = sin8x + sin4x ⇒ sin4x – sin2x = 0 ⇒ 2sin2x.cos2x – sin2x = 0 ⇒ sin2x (2cos2x – 1) = 0 ⇒ sin2x = 0 or 2cos2x – 1 = 0 2 1 ⇒ x = 2 nπ 3 π ⇒ x = nπ ± 6 π ∴ Solution of given equation is 2 nπ, n or nπ ± 6 π, n Ans. Type - 5
[PDF File]Subject : Mathematics Topic : TRIGONOMETRI EQUATIONS
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∵ sin5x.cos3x = sin6x.cos2x ⇒ 2sin5x.cos3x = 2sin6x.cos2x ⇒ sin8x + sin2x = sin8x + sin4x ⇒ sin4x – sin2x = 0 ⇒ 2sin2x.cos2x – sin2x = 0 ⇒ sin2x (2cos2x – 1) = 0 ⇒ sin2x = 0 or 2cos2x – 1 = 0 2 1 ⇒ x = 2 nπ 3 π ⇒ x = n π ± 6 π ∴ Solution of given equation is 2 nπππ, n or nπππ ± 6 πππ, n Ans. Type - 5
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sin2x 2cos2x 1 sinx 4cosx ... Áp dụng: 2sin 2x 3sin2x 1 sin2x 1 2sinx 12 ... cos2x cosx 1 cos3x cos3x cosx cos2x 1 0 ...
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