another cambridge question (1 Viewer)

[unmentionable]

yr11 book 14g
13c)
Show that d/dx (cosx + sinx)/(cosx - sinx) = sec^2(pi/4 + x)

thanks

 

[Mill]

d/dx (cosx + sinx)/(cosx - sinx)

= [ (cosx - sinx)(cosx + sinx) + (cosx + sinx)(cosx + sinx) ] / [ (cosx - sinx)^2 ]

= [ (cosx)^2 - 2sinxcosx + (sinx)^2 + (cosx)^2 + 2sinxcosx + (sinx)^2 ] / [ (cosx - sinx)^2 ]

= 2 / [ (cosx - sinx)^2 ]

= [ root(2) / (cosx - sinx) ]^2

= [ 1 / ( (cosx) / root(2) ) - ( (sinx) / root(2) ) ]^2

= [ 1 / ( (cosx)(cospi/4) - (sinx)(sinpi/4) ) ]^2

= [ 1 / (cos(pi/4 + x) ]^2

= sec^2(pi/4 + x)

 

[victorling]

the key is applying "rationalization"

 

[unmentionable]

thanks for the help
but how do u get from the first line to the second?

Originally posted by Mill


= [ root(2) / (cosx - sinx) ]^2

= [ 1 / ( (cosx) / root(2) ) - ( (sinx) / root(2) ) ]^2

did u just times top and bottom by 1/root2?

 

[KeypadSDM]

Originally posted by unmentionable
yr11 book 14g
13c)
Show that d/dx (cosx + sinx)/(cosx - sinx) = sec^2(pi/4 + x)

thanks

Note:
(cosx + sinx)/(cosx - sinx)
= (1 + sin2x)/(cos2x)
= sec2x + tan2x
d/dx (cosx + sinx)/(cosx - sinx)
= 2sec2xtan2x + 2(sec2x)^2
=2(1 + sin2x)/(cos2x)^2
=2(cosx + sinx)/[(cosx - sinx) * (cosx^2 - sinx^2)]
=2/(cosx - sinx)^2

Damn, I thought it was going to be quicker. Oh well.

 

[KeypadSDM]

/
|Sec^2[pi/4 + x]dx
/
=Tan[pi4 + x] + c
=(Tan[pi/4] + Tan[x])/(1 - Tan[pi/4]Tan[x]) + c
=(1 + Tan[x])/(1 - Tan[x]) + c
=(Cos[x] + Sin[x])/(Cos[x] - Sin[x]) + c
:. d/dx (Cos[x] + Sin[x])/(Cos[x] - Sin[x]) = Sec^2[pi/4 + x]

 

[Mill]

Yeah, top and bottom times 1 / root(2) but inside the brackets.