SOLUTION: If tan(x/2)=sinx and cosx≠0, then the value of tanx is…

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Question 1199347: If tan(x/2)=sinx and cosx≠0, then the value of tanx is…
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
tan%28x%2F2%29%22%22=%22%22sin%28x%29

One obvious solution is x=0%2Bn%2A2pi%22%22=%22%22x=n%2A2pi

sin%28x%2F2%29%2Fcos%28x%2F2%29%22%22=%22%22sin%282%2Aexpr%28x%2F2%29%29

sin%28x%2F2%29%2Fcos%28x%2F2%29%22%22=%22%222sin%28x%2F2%29cos%28x%2F2%29

Divide both sides by sin%28x%2F2%29, assuming x%3C%3E0, where
x=0 is true in the obvious solution above.

1%5E%22%22%2Fcos%28x%2F2%29%22%22=%22%222cos%28x%2F2%29

1%22%22=%22%222cos%5E2%28x%2F2%29

1%2F2%22%22=%22%22cos%5E2%28x%2F2%29

%22%22%2B-sqrt%281%2F2%29%22%22=%22%22cos%28x%2F2%29

cos%28x%2F2%29%22%22=%22%22%22%22%2B-sqrt%28%281%2A2%29%2F%282%2A2%29%29

cos%28x%2F2%29%22%22=%22%22%22%22%2B-sqrt%28%282%29%2F%284%29%29

cos%28x%2F2%29%22%22=%22%22%22%22%2B-sqrt%282%29%2F2

x%2F2%22%22=%22%22%22%22%2B-pi%2F4%2B2pi%2An

x%22%22=%22%22%22%22%2B-pi%2F2%2B4pi%2An

x%22%22=%22%224pi%2An%2B-pi%2F2

x%22%22=%22%22pi%2A%284n+%2B-+1%2F2+%29, where n is any integer.

and also the obvious solution x%22%22=%22%22n%2A2pi

Edwin


Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
If tan(x/2)=sin(x) and cos(x)≠0, then the value of tan(x) is . . .
~~~~~~~~~~~~~~~~~~

Our starting equation is 

    tan%28x%2F2%29 = sin%28x%29.      (1)


Since tan%28x%2F2%29 is not defined when cos%28x%2F2%29 = 0, the domain of this equation is 
the set of all real numbers except x%2F2 = pi%2F2%2Bk%2Api,  or  x = pi%2B2k%2Api = %282k%2B1%29%2Api.


One obvious solution to equation (1) is x = 2k%2Api, when both sides of equation (1) are zeroes.    (2)


In what follows, we assume that x is neither 2k%2Api nor %282k%2B1%29%2Api.    (3)


        In particular, it implies that sin%28x%2F2%29 =/= 0.    (4)


Now we transform equation (1) step by step

    sin%28x%2F2%29%2Fcos%28x%2F2%29 = sin%282%2A%28x%2F2%29%29

    sin%28x%2F2%29%2Fcos%28x%2F2%29 = 2sin%28x%2F2%29cos%28x%2F2%29


Due to (4), we can divide both sides by sin%28x%2F2%29, since this factor is not zero.

    1%2Fcos%28x%2F2%29 = 2cos%28x%2F2%29

    1 = 2cos%5E2%28x%2F2%29

    1%2F2 = cos%5E2%28x%2F2%29

    cos%28x%2F2%29 = %22%22%2B-sqrt%281%2F2%29 = %22%22%2B-sqrt%282%29%2F2

    x%2F2 = %22%22%2B-pi%2F4+%2B+pi%2An    <<<---=== notice here the difference with the Edwin's solution
                                 (here Edwin lost some/(half of the) solutions)


    x = %22%22%2B-pi%2F2+%2B+2pi%2An = 2pi%2An+%2B-+pi%2F2 = pi%2A%282n+%2B-+1%2F2+%29, where n is any integer.

              But for all these values of x, cos(x) = 0, so we should exclude this set of values
              from our consideration due to condition cos(x) =/= 0, imposed in the problem.


Finally, the only set of solutions is  x = 2k%2Api, found in (2).


ANSWER.  The solution to equation (1) with the imposed condition cos(x) =/= 0 

         is the set  x = 2k%2Api, and tan(x) = 0 for all these values of x.

Solved.