It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.
sin2x + cos x = 0
(1 - cos2x) + cos x = 0
-cos2x + cos x + 1 = 0
cos2x - cos x - 1 = 0
Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.
sin2x + cos x = 0
(1 - cos2x) + cos x = 0
-cos2x + cos x + 1 = 0
cos2x - cos x - 1 = 0
Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.
sin2x + cos x = 0
(1 - cos2x) + cos x = 0
-cos2x + cos x + 1 = 0
cos2x - cos x - 1 = 0
Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.
sin2x + cos x = 0
(1 - cos2x) + cos x = 0
-cos2x + cos x + 1 = 0
cos2x - cos x - 1 = 0
Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
cosx * tanx = sinx cos x . sin x/ cosx = sinx cos x . sin x/cosx = sinx (cos x is taken out of the equation, leaving sin x = sin x).
22
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.
No, (sinx)^2 + (cosx)^2=1 is though
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
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
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!
The differentiation of sin x plus cosx is cos (x)-sin(x).
(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)
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
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx