Question

The value of \[\sin {60^ \circ } - \cos {30^ \circ }\]
A) 0
B) \[\dfrac{1}{{\sqrt 2 }}\]
C) \[\dfrac{{\sqrt 3 }}{2}\]
D) 1

Answer 1

Hint: Apply the formula \[\sin (90 - x) = \cos x\] try to put \[x = 30\] and see if you can find any relation between \[\sin {60^ \circ }\& \cos {30^ \circ }\] use that relation to find the value of the given question
Complete step by Step solution:
We know that \[\sin (90 - x) = \cos x\] by applying this and putting the value of \[x = 30\] we are getting
\[\begin{array}{l}
\sin (90 - x) = \cos x\\
\sin (90 - 30) = \cos 30\\
\therefore \sin 60 = \cos 30
\end{array}\]
Now it is clear that \[\sin 60 = \cos 30\]
\[\therefore \sin {60^ \circ } - \cos {30^ \circ } = 0\]
Therefore, option A is correct.

Note: We can also do this sum by a different approach. First see the value of \[\sin {60^ \circ }\& \cos {30^ \circ }\]you will find that both is \[\dfrac{{\sqrt 3 }}{2}\] so if you just put there values we will get it as
\[\begin{array}{*{20}{l}}
{\therefore \sin {{60}^ \circ } - \cos {{30}^ \circ }}\\
{ = \dfrac{{\sqrt 3 }}{2} - \dfrac{{\sqrt 3 }}{2}}\\
{ = 0}
\end{array}\]
So, in both the cases we will get the same answer.