answersLogoWhite

0


Best Answer

0

User Avatar

Jadon Zulauf

Lvl 10
โˆ™ 2021-02-26 19:08:00
This answer is:
User Avatar
Study guides

Algebra

20 cards

A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

โžก๏ธ
See all cards
3.76
โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…
853 Reviews
More answers
User Avatar

Wiki User

โˆ™ 2013-02-23 06:04:20

You may have a typo in your question. I'll answer it for sinx/(1-cosx)

Multiply the numerator and denominator by (1 +cosx) and the denominator becomes (1-cos2x). By an identity, 1-cos2x = sin2x, giving you one term in the numerator.

Your expression simplifies to sinx(1+cosx)/sin2x. Factor out sinx and you get (1+cosx)/sinx.

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: How do you break 1 sinx divided 1-cosx?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

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


What is the derivative of 1 divided by sinx?

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


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


What is the anti-derivative of 4cosx divided by sinx to the power of 2?

2


How do you verify the identity sinx cscx 1?

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.


What is 1- sinx?

If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90


How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?

You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0


Is 1 plus sinX divided by 1 plus cscX equal to sinX?

2


What is the derivative of the square root of 1-sinx?

√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]


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 differentiate sine squared x?

Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)

People also asked