Saturday, June 29, 2013

Verify if the identity holds if x is in [0,pi/2]. (1+sinx)/(1+cosx)=(tan(x/2))'+tan(x/2)

We'll manage the right side of the expression and we'll
differentiate tan(x/2).


[tan(x/2)]' =
1/2*[cos(x/2)]^2


We'll re-write the
expression:


(1+sinx)/(1+cosx)=1/2*[cos(x/2)]^2 +
tan(x/2)


We'll replace the term tan (x/2) by the fraction
sin (x/2)/cos (x/2).


(1+sinx)/(1+cosx)=1/2*[cos(x/2)]^2 +
sin (x/2)/cos (x/2)


We'll multiply the term sin (x/2)/cos
(x/2) by 2*cos (x/2):


(1+sinx)/(1+cosx)=[1 + 2*cos
(x/2)*sin(x/2)]/2*[cos(x/2)]^2


We'll recognize the formula
of double angle:


2*cos (x/2)*sin(x/2) = sin 2*(x/2) = sin
x


(1+sinx)/(1+cosx)=(1+sinx)/2*[cos(x/2)]^2


We'll
replace the denominator from the right side by the formula of half
angle:


(1+sinx)/(1+cosx)=(1+sinx)/(1+cosx)


We
notice that managing RHS, we'll get LHS, therefore the identity is verified, if x
belongs to the range [0 , pi/2].

No comments:

Post a Comment

What accomplishments did Bill Clinton have as president?

Of course, Bill Clinton's presidency will be most clearly remembered for the fact that he was only the second president ever...