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d/dx 2 cos x = -2 sin x
cos2x/cosx = 2cosx - 1/cosx
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
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d/dx 2 cos x = -2 sin x
cos2x/cosx = 2cosx - 1/cosx
take out the constant -2 then take the intergral of cosx this will give you sinx your answer is -2sinx
30 degrees explanation 2Cosx-radical 3=0 Then 2cosx=radical 3 and cos x=(radical 3)/2 Now remember that cos 300 is (radical 3)/2 from the 30/60/90 triangle. So the answer is 30 degrees.
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees
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For f(x) + g(x), the derivative d/dx[f(x) + g(x)] = f'(x) + g'(x). Essentially, this just means that since it's addition, you can take the derivative of each part.d/dx(x - 2cosx) =* d/dx is a way to indicate you're taking the derivatived/dx(x) + d/dx(-2cosx)* take the derivative of each part, distributing(1) + (-2*-sinx)* f(x) = x, f'(x) = 1 & g(x) = -2cosx, g'(x) = +2sinx= 1 + 2sinx
To solve this equation, you must first put the equation in terms of cosx.3cosx=2cosx = 2/3Next, you find the reference angle (α) by finding the cos inverse of 2/3.α=cos-1(2/3) = 48.19 Degrees (approximately)find the distance from from 48.19 and 90 you then add the difference to 270 giving you the two answers which are ... 48.19 and 311.81 (approximately)
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x