Question (1 Viewer)

[j@son]

Show that d/dx[cosx+sinx/cosx-sinx]=sec^2(pi/4+x)

Thanx.

 

[Riviet]

Using quotient rule we obtain:

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

Expanding and simplifying this,

= 2(sin²x+cos²x) / (1-2sinxcosx)

= 2 / (1-sin2x)

= 2 / (1+sin(-2x)), since y=sinx is odd

= 1 / 1/2.(1+cos(pi/2+2x)), using identity sinA = cos(pi/2-A)

= 1 / cos²(pi/4+x), using identiy cos²A = 1/2.(1+cos2A)

= sec²(pi/4+x)

P.S Iruka's manipulation is very clever.

 

[icycloud]

Seems like they're asking for the auxiliary angle transformation.

Let y = (cosx+sinx)/(cosx-sinx) = P(x)/Q(x)
Notice that P(x) = sinx+cosx = Sqrt[2]sin(x+pi/4) and Q(x) = cosx-sinx = Sqrt[2]cos(x+pi/4) {by the auxiliary angle method}.

Thus, y = tan(x+pi/4), y' = sec²(x+pi/4), as required.

Note: Auxiliary angle method states that for R(x) = asinx+bcosx, we have R(x) = Sqrt[a^2+b^2]sin(x+arctan(b/a)) and S(x) = acosx-bsinx then S(x)=Sqrt[a^2+b^2]cos(x+arctan(b/a)). These results are easily proven.

 

[icycloud]

BTW, in these types of questions where they say "show that X = Y", you can always start with Y and show it's equal to X.

In this case, integrating RHS, we get:

Int(RHS) = tan[pi/4+x] + C
= (1+tanx)/(1-tanx) + C
= (cosx+sinx)/(cosx-sinx) + C {multiplying top and bottom by cosx}

Differentiating both sides, we have RHS = d/dx[LHS] + d/dx[C] = d/dx[LHS] as required.