SOLUTION: If (cos41° + sin41°)^2 = 2(sin^2 A), where 0° < A < 90°, what is the degree measure of angle A?

You are given

    (cos(41°) + sin(41°))^2 = 2(sin^2(A))      (1)



Square left side of this equation.   You will get

    
    1 + 2sin(41°)*cos(41°) = 2*sin^2(A),

or

    1 + sin(82°)           = 2sin^2(A).

    

Since  sin(82°) = cos(90°-82°) = cos(8°), you can re-write the last equation in an equivalent form

     = sin^2(A).                  (2) 



Now use the half-argument formula of Trigonometry 

     = cos^2(alpha/2}}}.


Then you can rewrite equation (2) in this form

     cos^2(4°) = sin^2(A).                     (3)



Next, recall that  cos(x) = sin(90°-x) for any angle x.  So, you can rewrite (3) as

     sin^2(86°) = sin^2(A).                    (4)



Since  0° < A < 90°, equation (4) implies

      A = 86°.



ANSWER.  A = 86°.