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Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
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
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
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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.
