Find the value angle A in degrees if $ \sin 2A = 1 $

Hint: Use different identities of trigonometric properties. Also, use the method of factorization for the quadratic equations. Use formulas like
 $
  {\sin ^2}\theta = 1 - {\cos ^2}\theta \\
  {(a - b)^2} = {a^2} - 2ab + {b^2} \\
  \sin 2A = 2\sin A\cos A \;
  $

Complete step-by-step answer:
Given that $ \sin 2A = 1 $
Substitute formula in the above equation, $ \sin 2A = 2\sin A\cos A $
 $
   \Rightarrow 2\sin A\cos A = 1 \\
  \therefore \sin A\cos A = \dfrac{1}{2} \\
  $
(Numerator in the multiplication with the terms goes in the denominator when changes its sides and vice –versa)
Also, substitute, $ \cos A = \sqrt {1 - {{\sin }^2}A} $
 $ \therefore \sin A(\sqrt {1 - {{\sin }^2}A} ) = \dfrac{1}{2} $
Let,
 $
  sinA = x \\
  \therefore si{n^2}A = {x^2} \\
  $
 $
   \Rightarrow x(\sqrt {1 - {x^2}} ) = \dfrac{1}{2} \\
   \Rightarrow \sqrt {1 - {x^2}} = \dfrac{1}{{2x}} \;
  $
Squaring on both the sides-
 $ $ $
   \Rightarrow {(\sqrt {1 - {x^2}} )^2} = {(\dfrac{1}{{2x}})^2} \\
   \Rightarrow 1 - {x^2} = \dfrac{1}{{4{x^2}}} \\
  $
[According to the property that squares and square root cancel each other]
Now, do the cross-multiplication and simplify-
 $
  \therefore 4{x^2}(1 - {x^2}) = 1 \\
  \therefore 4{x^2} - 4{x^4} = 1 \;
  $
Take all the terms on the left hand side –
 $ 4{x^2} - 4{x^4} - 1 = 0 $
Take negative sign common from all the sides –
 $ 4{x^4} - 4{x^2} + 1 = 0 $
Use perfect square formula in the above equation-
 $
  {(2{x^2} - 1)^2} = 0\;{\text{ [(a - b}}{{\text{)}}^2} = {a^2} - 2ab + {b^2}] \\
  \therefore (2{x^2} - 1) = 0 \\
  \therefore 2{x^2} = 1 \\
  \therefore {x^2} = \dfrac{1}{2} \\
  \therefore x = \dfrac{{ \pm 1}}{{\sqrt 2 }} \;
  $
Hence, put value of “x”
 $ sinA = \dfrac{1}{{\sqrt 2 }}{\text{ or sinA = }}\dfrac{{ - 1}}{{\sqrt 2 }} $
As, we know that –
 $
  sin45^\circ = \dfrac{1}{{\sqrt 2 }} \\
  sinA = sin45^\circ \\
  \therefore A = 45^\circ \\
  \therefore A = \dfrac{\pi }{4} \;
  $
 $ {\text{sinA = }}\dfrac{{ - 1}}{{\sqrt 2 }} $
Since Sin trigonometric functions are the odd functions
 $
  f( - x) = f(x) \\
  sinA = - sin45^\circ \\
  \therefore A = - 45^\circ \\
  \therefore A = \dfrac{{ - \pi }}{4} ;
  $
Hence, the required solution is –
The value of angle A is $ 45^\circ $ , if $ \sin 2A = 1 $
So, the correct answer is “ $ 45^\circ $ ”.

Note: Remember the properties of sines and cosines and apply accordingly. The odd and even trigonometric functions states that -
 $
  \sin ( - \theta ) = - \sin \theta \\
  \cos ( - \theta ) = \cos \theta \;
  $
The most important property of sines and cosines is that their values lie between minus one and plus one. Every point on the circle is unit circle from the origin. So, the coordinates of any point are within one of zero as well.
Directly the Pythagoras identity are followed by sines and cosines which concludes that –
 $ si{n^2}\theta + co{s^2}\theta = 1 $