Solve the equation $ \cos {57^ \circ } = \sin \boxed? $

Hint: Cosine and Sine functions are complementary to each other. Knowing this we will use the relation between the two functions and solve the equation
If $ x $ is an angle in degrees then,
 $ \cos x = \sin ({90^ \circ } - x) $

Complete step-by-step answer:
Cosine and Sine functions are complementary to each other.
Let $ \cos {57^ \circ } = \sin y $ --(1)
 $ \Rightarrow \sin ({90^ \circ } - {57^ \circ }) = \sin y $
 $ \Rightarrow ({90^ \circ } - {57^ \circ }) = y $ [ $ \because \sin a = \sin b \Rightarrow a = b $ ]
 $ \Rightarrow {33^ \circ } = y $ --(2)
Putting this in (1) we get:
 $ \cos {57^ \circ } = \sin {33^ \circ } $
Therefore, the solution to the given trigonometric problem is $ \cos {57^ \circ } = \sin {33^ \circ } $

Note: Since the given measurement of angle is in degrees don’t convert the angles to radians unnecessarily. Always remember that $ cos(90-x)=sinx $ and $ sin(90-x)=cosx $ where the measure of the angles is in degrees.