Sinx 2 3 2cosx

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      1 sinx+xcosx 0 2 2cosx−xsinx 2 3 −3sinx−xcosx 0 4 −4cosx+xsinx −4 5 5sinx+xcosx (a) f(x)=xsinx≈ T4(x)= 2 2! (x−0)2 + −4 4! (x−0)4 =x2 − x4 6 (b) |R4(x)| ≤ M 5! |x|5,where f(5)(x) ≤ M. Now −0.6 ≤ x ≤ 0.6 ⇒ |x| ≤ 0.6, and a graph of f(5)(x) shows that f(5)(x) ≤ 5for−0.6≤ x≤ 0.6. Thus, wecantakeM =4andget ...

    • [PDF File]3.2 ProvingIdentities - All-in-One High School

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      (1 sinx)2 1 sin2 x!be careful, these are NOT the same! Step 3: Factor the denominator and cancel out like terms. (1 sinx)2 (1+sinx)(1 sinx) 1 sinx 1+sinx 9.Plug in 5p 6 for x into the formula and simplify. 2sinxcosx =sin2x 2sin 5p 6 cos 5p 6 =sin2 5p 6 2 p 3 2! 1 2 =sin 5p 3 This is true because sin300 is p 3 2 10.Change everything into terms ...

    • SELVAM COLLEGE OF TECHNOLOGY,NAMAKKAL DEPARTMENT OF ECE TWO MARK - Vidyarthiplus

      3 2 sin 3 1 45.C.F for (y11 – 5y 1 + 6) y = 0 - c) –ex(xsinx + 2cosx) 46.C.F for (D 2 + 6D + 5) y = 0 - d) Acosx + Bsinx 47.P.I for (D 2 + 4) y = x sinx - e) Ae 2x + Be 3x 48.P.I for (D 2 – a2)y = e ax - f) + − 2 xsinx cosx sinx 49.P.I for (D 2 + 2D + 1)y = x sinx - g) (A + Bx)e ax 50.P.I for (D 2–2D+1)y = xe xsinx - h) Ae-x + Be-5x ...

    • [PDF File]HONORS PRECALCULUS Prove the following identities-

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      11. 2 cos 2 x =1 12. 4cos2 x −3= 0 Without a calculator compute the following: 13. sin 105° Lets sinx = -5/13 and cos y = 4/5 not in quadrant four find each of the following: 14. sin2x 15. tan 2x 16. cos (x+y) verify the following identities

    • [PDF File]Math 2250 Exam #2 Solutions

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      not horizontal at the origin, so the graph has a horizontal tangent line only when x = 2=3. Using the original equation of the curve to solve for the y-coordinates corresponding to x = 2=3, we see that y2 = ( 2=3)2( 2=3 + 1) = 4 9 1 3 = 4 27: Therefore, y = q 4 27 = 2 3 p 3. We conclude, then, that the Tschirnhausen cubic has a horizontal ...

    • [PDF File]FORMULARIO

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      Posto t = tan(x/2), si ha: sinx = 2t 1+t2; cosx = 1−t2 1+t2; tanx = 1−t2; sin0 = 0 cos0 = 1 sin π 6 = 1 2; cos π 6 = √ 3 2; sin 4 =

    • [PDF File]Assignment-4 - University of California, Berkeley

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      2e sinxcosx 2cosx+ f(x) 2e x 3!0 as x!1. This proves that lim x!1 2e sinxcosx 2cosx+ f(x) = 0: On the other hand, f(x) g(x) = e sinx which clearly does not have a limit as x!1. (d)Explain why this does not contradict L’Hospital’s rule. Solution: One of the assumptions when using L’Hospital’s rule when computing lim

    • [PDF File]USEFUL TRIGONOMETRIC IDENTITIES

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      sinx cotx= 1 tanx Fundamental trig identity (cosx)2 +(sinx)2 = 1 1+(tanx)2 = (secx)2 (cotx)2 +1 = (cosecx)2 Odd and even properties cos( x) = cos(x) sin( x) = sin(x) tan( x) = tan(x) Double angle formulas sin(2x) = 2sinxcosx cos(2x) = (cosx)2 (sinx)2 cos(2x) = 2(cosx)2 1 cos(2x) = 1 2(sinx)2 Half angle formulas sin(1 2 x) 2 = 1 2 (1 cosx) cos(1 ...

    • [PDF File]π 3π nπ 2 f π f 3π 2 π nπ, 3π nπ, n - WebAssign

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      ⇒ f′(x)=2cosx+2sinxcosx =0 ⇔ 2cosx(1+sinx)=0 ⇔ cosx =0 or sinx = −1, so x = π 2 +2nπ or 3π 2 +2nπ, where n is any integer. Now f π 2 = 3 and f 3π ...

    • [PDF File]STEP Support Programme STEP 2 Trigonometry Questions: Solutions - Maths

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      = (2cos2 x 1)cosx (2sinxcosx)sinx = 2cos3 x cosx 2cosxsin2 x = 2cos3 x cosx 2cosx(1 cos2 x) = 2cos3 x cosx 2cosx+ 2cos3 x = 4cos3 x 3cosx Since the answer is given, you do need to show every step. Remember \One equal sign per line, all equal signs aligned"! Similarly, using sin3x= sin2xcosx+ cos2xsinxleads to sin3x= 3sinx 4sin3 x.

    • [PDF File]MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS Section 2.6 (ODE Book)

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      2 MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS Moreover, we must have ex cosy +2cosx = N(x,y) = ψ y(x,y) = ex cosy +2cosx+c′(y). We must have that c′(y) = 0 and hence c(y) = c for some constant c.Therefore, the implicit form of the solution to the differential equation is

    • [PDF File]wp.kunduz.com

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      4cos2x—2 3 denkleminin kökler toplamr asaéldakilerden hangisi- dir? 12 511 C) 12 12 4 (2cosx—E . CDS X 2 11. x e [O,TC] olmak üzere = 4 cos x sinx— cosx sin x + cosx denkleminin çözüm kümesi asaéldakilerden hangisi olabilir? 2TC 3'2'3 3 2 511 6 C) 6'4'2 ( shx + k) 2 4 X e—sx —G 2 x ) 10. olmak üzere sin cos 2x = 1

    • [PDF File]Exercise 59

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      Stewart Calculus 8e: Section 3.4 - Exercise 59 Page 1 of 2 Exercise 59 Find all points on the graph of the function f(x) = 2sinx+sin2 xat which the tangent line is horizontal. Solution

    • [PDF File]14 Graphs of the Sine and Cosine Functions

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      2 ˇ 3ˇ 2 2ˇ y 2 0 -2 0 2 The graph is a vertical stretch by a factor of 2 of the graph of cosxas shown in Figure 14.7. Figure 14.7 Example 14.3 Sketch one cycle of the graph of y= cosˇx: Solution. The amplitude of the function is 1 and the period is 2ˇ b = 2ˇ ˇ = 2: x 0 1 2 1 3 2 2 y 1 0 -1 0 1 5

    • [PDF File]Section 5.3, Solving Trigonometric Equations

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      3) = 0 tanx= 0 tanx= p 3 x= nˇ x= ˇ 3 + nˇ 3.4sin2 x 1 = 0 4sin2 x 1 = 0 (2sinx 1)(2sinx+ 1) = 0 sinx= 1 2 sinx= 1 2 x= ˇ 6 + 2nˇ; 5ˇ 6 + 2nˇ x= 7ˇ 6 + 2nˇ; 11ˇ 6 + 2nˇ 4.3sin 2x= cos x Here, we have more than one trigonometric function, so we want to change this to having only one trigonometric function. Let’s use the identity ...

    • [PDF File]Trigonometric Identities - Miami

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      2 cos x y 2 sinx siny= 2sin x y 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

    • [PDF File]Quiz 3

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      (7 cos2 x) 2=3 ( 2cosx)( sinx) G0(x) = sin2x 3(7 cos2 x)2=3. 1. 1. (1 pm) [5 points] A graph of a function is given. In the coordinate plane below plot the graph of its derivative. Mark all important points. 2. 2. (12 pm) [5 points] Find the Slope-Intercept form of the equation of the tangent line to the

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