Question 1088217
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If  cos(A) - sin(A) = sqrt(2)*sin(A) then cos(A) + sin(A) equals
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<pre>
You are given

cos(A) - sin(A) = sqrt(2)*sin(A).    (1)


Divide both sides by {{{sqrt(2)}}}. You will get

{{{(1/sqrt(2))*cos(A)}}} - {{{(1/sqrt(2))*sin(A)}}} = sin(A).


It is the same as

{{{(sqrt(2)/2)*cos(A)}}} - {{{(sqrt(2)/2)*sin(A)}}} = sin(A).      (2)


Now recall that {{{sqrt(2)/2}}} = {{{sin(pi/4)}}} = {{{cos(pi/4)}}}.


Therefore, you can re-write (2) in the form

{{{sin(pi/4)*cos(A)}}} - {{{cos(pi/4)*sin(A)}}} = sin(A).


Using the adding/subtracting formula for sine, it is the same as

{{{sin(pi/4 - A)}}} = {{{sin(A)}}},                         (3)


which implies EITHER

    {{{pi/4 - A}}} = {{{A}}} + {{{2k*pi}}},                     (4)    

OR

    {{{pi/4 - A}}} + {{{A}}} = {{{pi + 2k*pi}}}                   (5)

where k is any integer.


Equation (5) has no solution, obviously.

Equation (4) has the solution

    2A = {{{pi/4+ 2k*pi}}},   or   A = {{{pi/8 + k*pi}}}.      (6)


Actually, we have two cases:  A = {{{pi/8}}}  and  A = {{{9pi/8}}}.


It is well known fact that 

{{{sin(pi/8)}}} = {{{sqrt(2-sqrt(2))/2}}},  {{{cos(pi/8)}}} = {{{sqrt(2+sqrt(2))/2}}}.

    (see the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Miscellaneous-Trigonometry-problems.lesson>Miscellaneous Trigonometry problems</A> in this site).


So, if A = {{{pi/8}}},   then  cos(A) + sin(A) = {{{sqrt(2+sqrt(2))/2}}} + {{{sqrt(2-sqrt(2))/2}}}.


    If A = {{{9pi/8}}},  then  cos(A) + sin(A) = -( {{{sqrt(2+sqrt(2))/2}}} + {{{sqrt(2-sqrt(2))/2}}} ).
</pre>

<U>Answer</U>.  &nbsp;If  &nbsp;cos(A) - sin(A) = sqrt(2)*sin(A) &nbsp;&nbsp;then 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;a)  &nbsp;A = {{{pi/8}}}  &nbsp;or  &nbsp;A = {{{9pi/8}}},  &nbsp;&nbsp;and


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b) &nbsp;cos(A) + sin(A) &nbsp;equals &nbsp;&nbsp;{{{sqrt(2+sqrt(2))/2}}} + {{{sqrt(2-sqrt(2))/2}}}  &nbsp;&nbsp;or  &nbsp;&nbsp;-({{{sqrt(2+sqrt(2))/2}}} + {{{sqrt(2-sqrt(2))/2}}}).


Solved.