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You can transform this in different ways (I suggest writing cot x as "cos x / sin x", for easier manipulation), but I doubt you can make it any simpler.
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Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
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
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
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.
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
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
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
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 integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
What is sinx plus cosx plus cotx simplified?
There is no sensible or useful simplification.
How do you solve sin x - 2 sin x is equal to 0?
sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...
Find the critical numbers of sinx plus cosx?
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
Integral of cos2x log cosx-sinx coax plus since?
it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.
Verify that Cos theta cot theta plus sin theta equals csc theta?
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)
