Question

Prove that \[\operatorname{Sin} {38^ \circ } + \operatorname{Sin} {22^ \circ } = \operatorname{Sin} {82^ \circ }\]

Answer 1

Hint: Here we will solve the left hand side of the given equation and get to right hand side
We can use the following identities to solve the LHS
\[SinA + SinB = 2Sin(\dfrac{{A + B}}{2})\operatorname{Cos} (\dfrac{{A + B}}{2})\]
\[\cos \left( {90 - \theta } \right) = \sin \theta \]

Complete step-by-step answer:
Start solving with L.H.S
\[LHS = \operatorname{Sin} {38^ \circ } + \operatorname{Sin} {22^ \circ }\]
Applying the following identity:
$\operatorname{Sin} A + \operatorname{Sin} B = 2\operatorname{Sin} (\dfrac{{A + B}}{2})Cos(\dfrac{{A + B}}{2})$
We get:-
Since in LHS
\[A = {38^ \circ }\]and \[B = {22^ \circ }\]
Putting in the values in the identity we get:-
$ \Rightarrow $\[\operatorname{Sin} {38^ \circ } + \operatorname{Sin} {22^ \circ } = 2\operatorname{Sin} \left( {\dfrac{{38 + 22}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{38 - 22}}{2}} \right)\]
After adding and subtracting the angles we get
$ \Rightarrow $\[LHS = 2\operatorname{Sin} \left( {\dfrac{{60}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{16}}{2}} \right)\]
Dividing the angles by 2 we get:-
$ \Rightarrow $\[LHS = 2\operatorname{Sin} ({30^ \circ })\operatorname{Cos} {8^ \circ }\]
Now we know that,
The value of \[\sin ({30^ \circ }) = \dfrac{1}{2}\]
Hence putting this value in above equation we get:-
$ \Rightarrow $\[LHS = 2.\dfrac{1}{2}.\operatorname{Cos} {8^ \circ }\]
$ \Rightarrow $\[LHS = \operatorname{Cos} {8^ \circ }\]
Now since we know that
\[\cos \left( {90 - \theta } \right) = \sin \theta \]
Hence applying this identity we get:-
$ \Rightarrow $\[\cos {8^ \circ } = \cos \left( {{{90}^ \circ } - {{82}^ \circ }} \right)\]
 \[ = \sin {82^ \circ }\]
Hence \[LHS = RHS\]
Hence proved.

Note: When you don’t have the perfect angle in these types of questions you should use identities which have sum of angles and start solving by taking one side which is RHS or LHS using the identities which have sum of angles.