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Question: How do you find the exact functional value of \(\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \ci...

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

Explanation

Solution

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:
⇒sin⁡C+sin⁡D=2sin⁡(C+D2)cos⁡(C−D2)\Rightarrow \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)
⇒sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\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:
⇒sin⁡C+sin⁡D=2sin⁡(C+D2)cos⁡(C−D2)\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:
⇒12(sin⁡C+sin⁡D)=sin⁡(C+D2)cos⁡(C−D2)\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:
⇒sin⁡(C+D2)cos⁡(C−D2)=12(sin⁡C+sin⁡D)\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=2cos⁡(C+D2)sin⁡(C−D2)\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right), as shown below:
⇒cos⁡(C+D2)sin⁡(C−D2)=12(sin⁡C−sin⁡D)\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∘cos⁡20∘−cos⁡110∘sin⁡20∘\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }, now consider the first part of the expression, as given below:
⇒sin⁡110∘cos⁡20∘\Rightarrow \sin {110^ \circ }\cos {20^ \circ }
The above expression is in the form of sin⁡(C+D2)cos⁡(C−D2)\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right), on comparing :
Here C+D2=110∘\dfrac{{C + D}}{2} = {110^ \circ } and C−D2=20∘\dfrac{{C - D}}{2} = {20^ \circ }
Which gives C+D=220∘C + D = {220^ \circ } and C−D=40∘C - D = {40^ \circ }, solving these two equations gives:
Hence C=130∘C = {130^ \circ } and D=90∘D = {90^ \circ }
Now applying the trigonometric product to sum identity to sin⁡110∘cos⁡20∘\sin {110^ \circ }\cos {20^ \circ }, as given below:
∴sin⁡110∘cos⁡20∘=12(sin⁡130∘+sin⁡90∘)\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∘cos⁡20∘−cos⁡110∘sin⁡20∘\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }, as given below:
⇒cos⁡110∘sin⁡20∘\Rightarrow \cos {110^ \circ }\sin {20^ \circ }
∴cos⁡110∘sin⁡20∘=12(sin⁡130∘−sin⁡90∘)\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∘cos⁡20∘−cos⁡110∘sin⁡20∘\sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ }, as given below:
⇒sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=12(sin⁡130∘+sin⁡90∘)−12(sin⁡130∘−sin⁡90∘)\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)
⇒sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=12(sin⁡130∘+sin⁡90∘−sin⁡130∘+sin⁡90∘)\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∘\sin {130^ \circ } gets cancelled on the right hand side of the above equation, as given below:
⇒sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=12(sin⁡90∘+sin⁡90∘)\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∘=1\sin {90^ \circ } = 1, hence substituting it in the above equation:
⇒sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=12(1+1)\Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( {1 + 1} \right)
⇒sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=12(2)\Rightarrow \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = \dfrac{1}{2}\left( 2 \right)
∴sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘=1\therefore \sin {110^ \circ }\cos {20^ \circ } - \cos {110^ \circ }\sin {20^ \circ } = 1

The value of sin⁡110∘cos⁡20∘−cos⁡110∘sin⁡20∘\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:
⇒sin⁡C+sin⁡D=2sin⁡(C+D2)cos⁡(C−D2)\Rightarrow \sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})
⇒sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\Rightarrow \sin C - \sin D = 2\cos (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})
⇒cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\Rightarrow \cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})
⇒cos⁡C−cos⁡D=−2sin⁡(C+D2)sin⁡(C−D2)\Rightarrow \cos C - \cos D = - 2\sin (\dfrac{{C + D}}{2})\sin (\dfrac{{C - D}}{2})