2cos 2x cos2x 2

• [PDF File]WZORY TRYGONOMETRYCZNE - UTP

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  WZORY TRYGONOMETRYCZNE tgx = sinx cosx ctgx = cosx sinx sin2x = 2sinxcosx cos2x = cos 2x−sin x sin2 x = 1−cos2x 2 cos2 x = 1+cos2x 2 sin2 x+cos2 x = 1 ASYMPTOTY UKOŚNE y = mx+n m = lim x→±∞ f(x) x, n = lim x→±∞ [f(x)−mx]POCHODNE [f(x)+g(x)]0= f0(x)+g0(x)[f(x)−g(x)]0= f0(x)−g0(x)[cf(x)]0= cf0(x), gdzie c ∈R[f(x)g(x)]0= f0(x)g(x)+f(x)g0(x)h f(x) g(x) i 0 = f0(x)g(x)−f(x ...

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• [PDF File]Math 142, Quiz 2. 9/2/10. Name

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  1 cos2x 2 1+cos2x 2 dx = 1 4 Z ˇ=2 0 (1 2cos 2x)dx= 1 4 Z ˇ=2 0 1 1+cos4x 2 dx = 1 4 Z ˇ=2 0 dx 1 8 Z ˇ=2 0 dx 1 8 Z ˇ=2 0 cos4xdx = 1 4 ˇ 2 1 8 ˇ 2 1 8 1 4 sin4x ˇ=2 0 = ˇ 8 ˇ 16 (0 0) = ˇ 16: Note: Since we have even powers of both sin and cos, we used identities to manipulate the integral: (1) sin2x= 2sinxcosx =)sin2 2x= 4sin ...

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• [PDF File]Solutions: Section 2 - Whitman People

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  2cos(2x) 3+2y ⇒ (3+2y)dy = 2cos(2x)dx Integrate both sides, and use the initial condition y(0) = −1 3y +y2 = sin(2x)+C ⇒ −3+1 = 0+C ⇒ C = −2 The implicit solution is: y2 +3y = sin(2x)−2 We can solve this for y by completing the square: y2 +3y = y2 +3y + 9 4 − 9 4 = y + 3 2 2 − 9 4 so that: y + 3 2 2 = sin(2x)+ 1 4 ⇒ y = − ...

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• [PDF File]Trigonometry and Complex Numbers - Youth Conway

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  cos2x cos 2014ˇ2 x = cos4x 1: Solution. We see cos2xmultiple times on the left side, so this motivates us to write the right side as a function of cos2xwith the double angle identity. 2cos2x cos2x cos 2014ˇ2 x = cos4x 1 = 2cos2 2x 2: Now, we can divide by 2 and expand the left side. cos2 2x cos2xcos 2014ˇ2 x = cos2 2x 1: cos2xcos 2014ˇ2 x = 1:

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• [PDF File]6.2 Trigonometric Integrals and Substitutions

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  [cos2 x]2dx= Z 1 + cos2x 2 2 dx = 1 4 Z (1 + 2cos2x+ cos2 2x)dx = 1 4 Z dx+ 2 cos2xdx+ 1 + cos4x 2 dx = 1 4 x+ 2 sin2x 2 + x 2 + 1 2 sin4x 4 = 1 4 3 2 x+ sin2x+ sin4x 8 Example 6.2.3 Find R sin2 xdx: Solution. Using the trigonometric identity sin2 x= 1 cos2x 2 we nd Z sin2 xdx= Z 1 cos2x 2 dx = 1 2 Z (1 cos2x)dx = 1 2 x sin2x 2 + C Integrals of ...

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• [PDF File]DOUBLE-ANGLE, POWER-REDUCING, AND HALF-ANGLE FORMULAS

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  cos 2x = cos2 x – sin2 x = 1 – 2 sin2 x = 2 cos2 x – 1 • Tangent: tan 2x = 2 tan x/1- tan2 x = 2 cot x/ cot2 x -1 = 2/cot x – tan x . tangent double-angle identity can be accomplished by applying the same . methods, instead use the sum identity for tangent, first. • Note: sin 2x ≠ 2 sin x; cos 2x ≠ 2 cos x; tan 2x ≠ 2 tan x ...

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• [PDF File]Even powers (half-angle identity) - University of Washington

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  2 (1+cos(2x))]2 dx = 1 4 Z 1+2cos(2x)+cos2(2x)dx = 1 4 x+ 1 4 sin(2x)+ 1 4 Z cos2(2x)dx = 1 4 x+ 1 4 sin(2x)+ 1 8 Z 1+cos(4x)dx = 1 4 x+ 1 4 sin(2x)+ 1 8 x+ 1 32 sin(4x)+C = 3 8 x+ 1 4 sin(2x)+ 1 32 sin(4x)+C You can do sin4(x) and sin 2(x)cos (x) is a similar way as above. Odd powers (identity then substitution): Z cos3(x)dx = Z

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• [PDF File]Calculus II - Exam 2 - Techniques of Integration

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  1 cos2x 2 dx 1. 2. R e 1 x 3 lnxdx 2. 3. R e 2xsin(2x)dx 3. 4. R 2x+1 x2 7x+12 dx 4. 5. R x4 x2 1 dx 5. 6. R 35sin4 xcos3 xdx 6. 7. R 1=3 0 p 1 9t2dt 7. Determine if the following integral converges. 8. R 2 1 2 ... = cos2 x sin2 x = 2cos2 x 1 = 1 2sin2 x tan(2x) = 2tanx 1 2tan x Half-angle formulas sin2 x = 1 cos(2x) 2 cos2 x = 1+cos(2x) 2 ...

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• [PDF File]Trigonometry Identities I Introduction - Math Plane

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  2 sin sm 30 or 2) y = cos2x and y = cosx+2 (Set equations equal to each other) cos2x — cosx + 2 o Substitution (Double Angle Identity) Set equation equal to zero Re-arrange the polynomial F actor Solve 2cos x 2cos x 2cos x (2cosx 2cosx cosx 1 — cosx + 2 1 — cosx 2 cosx — 3 — 0 3)(cosx + 1) 0 1 cosx+ 1 0 cosx No Solution! (cost 1) COST ...

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• [PDF File]Solution 1. Solution 2.

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  cos2x+ cos2y cos2x )cos2y = 2cos(2x+2y 2)cos(2x 2y) 2sin(2x+2y 2)sin(2x y 2 = cot(x+ y)cot(x y) Solution 23. 6. Since a= 1 2 and b= 21 2 we nd k = q (1 2) + (1 2) 2= p 2;cos = a k = p 2 2;sin = b k = p 2 2. Thus 45 and y= p 2 2 sin(x 45 ): Solution 24.

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• [PDF File]Lecture 9 : Trigonometric Integrals Extra Examples Z cos xdx

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  (sin2 x)2cos 2xdx = Z [1 2 (1 cos2x)] [1 2 (1+cos2x)]dx = 1 8 Z (1 cos(2x))2(1+cos(2x))dx = 1 8 Z (1 cos2(2x))(1 cos(2x))dx you can deal with this in two ways number 1: = 1 8 Z sin2(2x)(1 cos(2x))dx = 1 8 [Z sin2(2x)dx Z sin2(2x)cos(2x))dx] = 1 8 [Z 1 2 (1 cos(4x))dx 1 2 Z sin2(w)cos(w))dw] 1. where w = 2x and dw = 2dx. = 1 16 [x 2 sin(4x) 4] 1 ...

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• [PDF File]Trig Equations with Half Angles and Multiple Angles angle

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  Example : Solve cos2x 3 2 over the interval 0,2 . Solution : Write the interval 0,2 as the inequality 0 x 2 and then multiply by 2 to obtain the interval for 2x: 0 2x 4 Using radian measure we find all numbers in this interval with cosine value 3 2. These are 6, 11 6, 13 6, and 23 6. So 2x 6, 11 6, 13 6, 23 6 x 12, 11 12, 13 12, 23 12 Write the ...

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• [PDF File]7.2 Trigonometric Integrals

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  7.2 Trigonometric Integrals The three identities sin 2x +cos x = 1, cos x = 1 2 (cos2x +1) and sin2x = 1 2 (1 cos2x) can be used to integrate expressions involving powers of Sine and Cosine.

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• [PDF File]Trigonometric Identities - Miami

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  2 cos x+y 2 cosx+ cosy= 2cos x+y 2 cos x y 2 cosx cosy= 2sin x+y 2 sin x y 2 The Law of Sines sinA a = sinB b = sinC c Suppose you are given two sides, a;band the angle Aopposite the side A. The height of the triangle is h= bsinA. Then 1.If a

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• [PDF File]Section 7.2 Advanced Integration Techniques: Trigonometric ...

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  2! 1 2cos(2x) + cos2(2x) 4! dx = 1 8 Z 1 2cos(2x) + cos2(2x) + cos(2x) 2cos2(2x) + cos3(2x) dx = 1 8 Z 1 cos(2x) cos2(2x) + cos3(2x) dx = 1 8 x 1 2 sin2x Z cos2(2x) dx+ Z cos3(2x)! dx: We have already showed that Z cos2(2x) dx= 1 2 x+ 1 8 sin(4x) + C and Z cos3(2x) dx= 1 2 sin(2x) 1 6 sin3(2x) + C; so nally we have Z cos2 xsin4 xdx= 1 8 x 1 2 ...

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• [PDF File]Math 113 Lecture #9 x7.2: Trigonometric Integrals Evaluate ...

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  1 + cos2x 8 dx = 1 8 Z 1 + cos2x 2cos2x 2cos2 2x+ cos2 2x+ cos3 3x dx = 1 8 Z 1 cos2x cos2 2x+ cos3 2x dx = 1 8 Z 1 cos2x dx 1 8 Z cos2 2xdx+ 1 8 Z cos3 2xdx: The rst integral is easy to compute, but how about the second and third integrals? We apply the half-angle formula to the second integral, and apply Case 1 followed by the

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• [PDF File]cos x cos x cos x x cos x x ...

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  2 sinx cosx sinxcosx 11 sin2x sin2x 22 .: ةصلاخ sin3x cos3x 2 sinx cosx .: نأ نيبن .2 sin2x sin4x sin6x 2sin2x 1 cos2x cos4x : انيدل 2 2 2x 6x 2x 6x 2sin cos sin4x sin2x sin4x sin6x sin2x sin6x sin4x 22 1 cos2x cos4x 1 2cos 2x 1 cos2x1 cos 2 2x cos2x 2sin 4x cos 2x sin4x sin 4x 2cos 2x 1 2cos 2x cos2x ...

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• [PDF File]Trigonometric Integrals{Solutions

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  (1+cos2x)=2 = cos2 x 4. sin(a)sin(b) = 1 2 [cos(a b) cos(a+b)] cos(a b) cos(a+b) = cosacosb+sinasinb (cosacosb sinasinb) cos(a b) cos(a+b) = 2sinasinb 1 2 [cos(a b) cos(a+b)] = sinasinb Integrals Evaluate the following integrals: 1. R sin2(p x)= xdx sub u = p x, then use the cosine substitution for sin2 u to get p x sin(2 p x)=2 2. Rp 1+cos(2x ...

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• [PDF File]Trigonometric Identities

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  cotn 2(x)dx Z secn(x)dx = tan(x)secn 2(x) n 1 + n 2 n 1 Z secn 2(x)dx Z cscn(x)dx = cot(x)cscn 2(x) n 1 + n 2 n 1 Z cscn 2(x)dx Other Integration Formulas Z dx x2 +a = 1 p a arctan x p a +C (for a > 0) Important Power Series 1 1 x = X1 k=0 xk = 1+x+x2 +x3 +::: ex = X1 k=0 xk k! = 1+x+ x2 2 + x3 6 +::: sin(x) = X1 k=0 ( k1) x2k+1 (2k +1)! = x x3 ...

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