Question

Prove that \[\sin {60^ \circ }.\cos {30^ \circ } - \cos {60^ \circ }.\sin {30^ \circ } = \sin {30^ \circ }\]

Answer 1

Hint: This question can be solved in more than one way. One way is to directly solve the question by comparing the equation by the formula of \[\sin ({\text{A}} - {\text{B}})\]. For this question, first of all we will compare the values of \[\sin ({\text{A}} - {\text{B}})\] to the equation given in the question, and just solve the question, we will get the correct answer.

Complete step by step answer:
In order to prove the given result, we have to start by considering the LHS first. Then, we will proceed and show that it is equal to the RHS.
As we know the formula of \[\sin ({\text{A}} - {\text{B}})\] is given by
\[\sin ({\text{A}} - {\text{B}}) = \sin {\text{A}}\cos {\text{B}} - \cos {\text{A}}\sin {\text{B}}\]
Now, comparing the values of the above equation to the values of equation given in the question, we get that values of A and B are
\[
  {\text{A}} = {60^ \circ } \\
  {\text{B}} = {30^ \circ } \\
\]
Hence putting the values of A and B in the above relation and simplifying further, we have
\[
  \sin ({\text{A}} - {\text{B}}) = \sin {\text{A}}\cos {\text{B}} - \cos {\text{A}}\sin {\text{B}} \\
  \sin ({60^ \circ } - {30^ \circ }) = \sin {60^ \circ }\cos {30^ \circ } - \cos {60^ \circ }\sin {30^ \circ } \\
  \sin {30^ \circ } = \sin {60^ \circ }\cos {30^ \circ } - \cos {60^ \circ }\sin {30^ \circ } \\
\]
Since we have got the LHS = RHS, we have proved the result.
Hence proved.

Note: We can solve the same question in another way by putting the respective numerical values. Taking the LHS of the equation and substituting the values, we have
\[
  \sin {60^ \circ }.\cos {30^ \circ } - \cos {60^ \circ }.\sin {30^ \circ } \\
   = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{1}{2} \\
   = \dfrac{3}{4} - \dfrac{1}{4} \\
   = \dfrac{2}{4} \\
   = \dfrac{1}{2} \\
\]
Now taking RHS of the same equation and substituting the values, we have
\[\sin {30^ \circ } = \dfrac{1}{2}\]
Therefore, \[{\text{LHS}} = {\text{RHS}}\]
Hence Proved.
For solving this question, we must remember the values of different trigonometric functions with respect to different values correctly. We can fail to solve the question correctly, if we forget or remember the wrong value of trigonometric functions. Also, for the first method, we must remember the correct formula and its comparing values. Other than this, we must know the basics of trigonometry so that we can put the values and solve the steps correctly.