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sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x)

sin2x=1-cos2x as sin2x+cos2x=1

1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)

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โˆ™ 2011-06-06 22:23:02
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During the 1930s the new deal legislation was proposed by president Franklin Roosevelt to bring the US out of the great depression

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Q: How does sin2x divided by 1-cosx equal 1 plus cosx?
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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


Does cos2x equal 2cosxsinx?

No; sin2x = 2 cosx sinx


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...:))


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


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How do you break 1 sinx divided 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.


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


Tan plus cot divided by tan equals csc squared?

(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.


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