Question

How do you find the exact functional value of $\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }$ using the cosine sum or difference identity?

Answer 1

Hint: This problem deals with trigonometric sum to product identities. Given an equation, we have to find the number of solutions of the equation. Which means that there are several solutions of the equation and hence we have to find the general solution of the equation. The formula used here is the trigonometric sum to product formula, which is given by:
$ \Rightarrow \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
$ \Rightarrow \sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$

Complete step-by-step solution:
From the above trigonometric sum to product formulas, we can rearrange the formulas in such a way that as shown below:
$ \Rightarrow \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
From here dividing the above equation by 2, gives:
$ \Rightarrow \dfrac{1}{2}\left( {\sin C + \sin D} \right) = \sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
So the trigonometric product to sum identity becomes, as shown below:
$ \Rightarrow \sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) = \dfrac{1}{2}\left( {\sin C + \sin D} \right)$
Similarly for the identity $\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$, as shown below:
$ \Rightarrow \cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right) = \dfrac{1}{2}\left( {\sin C - \sin D} \right)$
Given $\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }$, now consider the first part of the expression, as given below:
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ }$
The above expression is in the form of $\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$, on comparing :
Here $\dfrac{{C + D}}{2} = {110^ \circ }$ and $\dfrac{{C - D}}{2} = {20^ \circ }$
Which gives $C + D = {220^ \circ }$ and $C - D = {40^ \circ }$, solving these two equations gives:
Hence $C = {130^ \circ }$ and $D = {90^ \circ }$
Now applying the trigonometric product to sum identity to $\sin {110^ \circ }\cos {20^ \circ }$, as given below:
$\therefore \sin {110^ \circ }\cos {20^ \circ } = \dfrac{1}{2}\left( {\sin {{130}^ \circ } + \sin {{90}^ \circ }} \right)$
Now consider the second part of the expression of $\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }$, as given below:
$ \Rightarrow \cos {110^ \circ }\sin {20^ \circ }$
$\therefore \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {\sin {{130}^ \circ } - \sin {{90}^ \circ }} \right)$
Now consider the given expression $\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }$, as given below:
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {\sin {{130}^ \circ } + \sin {{90}^ \circ }} \right) - \dfrac{1}{2}\left( {\sin {{130}^ \circ } - \sin {{90}^ \circ }} \right)$
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {\sin {{130}^ \circ } + \sin {{90}^ \circ } - \sin {{130}^ \circ } + \sin {{90}^ \circ }} \right)$
Here $\sin {130^ \circ }$ gets cancelled on the right hand side of the above equation, as given below:
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {\sin {{90}^ \circ } + \sin {{90}^ \circ }} \right)$
We know that $\sin {90^ \circ } = 1$, hence substituting it in the above equation:
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {1 + 1} \right)$
$ \Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( 2 \right)$
$\therefore \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = 1$

The value of $\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }$ is 1.

Note: While solving this problem, we need to understand that throughout the problem we used one of the trigonometric sum to product formula, there are totally four such trigonometric sum to product formulas, which are given by:
\[ \Rightarrow \sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \sin C - \sin D = 2\cos (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})\]
\[ \Rightarrow \cos C - \cos D = - 2\sin (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})\]