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Euler's formula, e(ix) cos(x) isin(x), is proven using Taylor series expansion and complex numbers. By substituting the Taylor series expansions for e(ix), cos(x), and sin(x) into the formula and simplifying, it can be shown that the equation holds true for all real numbers x.
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How do you express sin 3x as a polynom with sin x using the De Moivre formula?
According to de Moivre's formula, cos3x + isin3x = (cosx + isinx)3 = cos3x + 3cos2x*isinx + 3cosx*i2sin2x + i3sin3x Comparing the imaginary parts, isin3x = 3cos2x*isinx + i3sin3x so that sin3x = 3cos2x*sinx - sin3x = 3*(1-sin2x)sinx - sin3x = 3sinx - 4sin3x
When did euler show that e raised to pi times imagenary number plus one equals zero?
The answer to this question is more complicated than might appear. First, Euler's formula, eix = cosx + i*sinx was known before Euler. For example Cotes discovered that ln(cosx + isinx) = ix. Taking natural antilogs gives Euler's formula. Cotes published in 1714 when Euler was aged only 7. Second, there is no record that shows that Euler simplified his formula and derived the identity that bears his name. Having said all that, Euler "discovered" the formula in 1740 and published its proof in 1748. Incidentally, I consider it to be the most beautiful formula EVER.
How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
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How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(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
Can you Show 1 over sinx cosx - cosx over sinx equals tanx?
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
How do you find the solutions of tanx equals 2cscx?
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
How do you solve sin squared x divided by 1 - cos x?
Use this identity sin2x+cos2x=1 sin2x=1-cos2x so sin2x/(1-cosx) =(1-cos2x)/(1-cosx) =(1-cosx)(1+cosx)/(1-cosx) =1+cosx
What is cosx times cosx?
It is cos2x that is, "cos-squared x".
What is the derivative of sin x minus cos x?
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
How do you Prove sin x times sec x equals tan x?
Remember SecX = 1/CosX Substitute SinX X 1 /CosX = SinX / CosX = TanX
