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Question: Solve the equation \( \cos {57^ \circ } = \sin \boxed? \)...

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

Explanation

Solution

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 xx is an angle in degrees then,
cos⁡x=sin⁡(90∘−x)\cos x = \sin ({90^ \circ } - x)

Complete step-by-step answer :
Cosine and Sine functions are complementary to each other.
Let cos⁡57∘=sin⁡y\cos {57^ \circ } = \sin y --(1)
⇒sin⁡(90∘−57∘)=sin⁡y\Rightarrow \sin ({90^ \circ } - {57^ \circ }) = \sin y
⇒(90∘−57∘)=y\Rightarrow ({90^ \circ } - {57^ \circ }) = y [ ∵sin⁡a=sin⁡b⇒a=b\because \sin a = \sin b \Rightarrow a = b ]
⇒33∘=y\Rightarrow {33^ \circ } = y --(2)
Putting this in (1) we get:
cos⁡57∘=sin⁡33∘\cos {57^ \circ } = \sin {33^ \circ }
Therefore, the solution to the given trigonometric problem is cos⁡57∘=sin⁡33∘\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)=sinxcos(90-x)=sinx and sin(90−x)=cosxsin(90-x)=cosx where the measure of the angles is in degrees.