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Question: How do you use the sum and difference formula to simplify \(\sin {20^ \circ }\cos {80^ \circ } - \co...

How do you use the sum and difference formula to simplify sin⁡20∘cos⁡80∘−cos⁡20∘sin⁡80∘\sin {20^ \circ }\cos {80^ \circ } - \cos {20^ \circ }\sin {80^ \circ } ?

Explanation

Solution

Hint : The given question deals with basic simplification of trigonometric functions by using some of the special trigonometric addition and subtraction formulae such as 2sin⁡Acos⁡B=sin⁡(A+B)−sin⁡(A−B)2\sin A\cos B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right) and 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right) . Basic algebraic rules and trigonometric identities are also to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer :
In the given problem, we have to simplify the trigonometric expression sin⁡20∘cos⁡80∘−cos⁡20∘sin⁡80∘\sin {20^ \circ }\cos {80^ \circ } - \cos {20^ \circ }\sin {80^ \circ } using use sum and difference formula.
So, sin⁡20∘cos⁡80∘−cos⁡20∘sin⁡80∘\sin {20^ \circ }\cos {80^ \circ } - \cos {20^ \circ }\sin {80^ \circ }
Multiplying and dividing the expression given to us by 22, we get,
⇒12(2sin⁡20∘cos⁡80∘−2cos⁡20∘sin⁡80∘)\Rightarrow \dfrac{1}{2}\left( {2\sin {{20}^ \circ }\cos {{80}^ \circ } - 2\cos {{20}^ \circ }\sin {{80}^ \circ }} \right)
Now, both the terms inside the bracket resemble the sum and difference formulae of trigonometry such as 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right) and 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right).
So, using the sum and difference formulae of trigonometry and doing a bit of simplification, we get,
⇒12(sin⁡(100)∘+sin⁡(−60∘)−sin⁡(100∘)+sin⁡(−60∘))\Rightarrow \dfrac{1}{2}\left( {\sin {{\left( {100} \right)}^ \circ } + \sin \left( { -{{60}^ \circ }} \right) - \sin \left( {{{100}^ \circ }} \right) + \sin \left( -{{{60}^ \circ }} \right)} \right)
⇒−12(sin⁡(60∘)+sin⁡(60∘))\Rightarrow -\dfrac{1}{2}\left( {\sin \left( {{{60}^ \circ }} \right) + \sin \left( {{{60}^ \circ }} \right)} \right)
⇒12(−2sin⁡(60∘))\Rightarrow \dfrac{1}{2}\left( {-2\sin \left( {{{60}^ \circ }} \right)} \right)
⇒sin⁡60∘\Rightarrow \sin {60^ \circ }
⇒(−32)\Rightarrow \left( {-\dfrac{{\sqrt 3 }}{2}} \right)
So, we get the value of sin⁡20∘cos⁡80∘−cos⁡20∘sin⁡80∘\sin {20^ \circ }\cos {80^ \circ } - \cos {20^ \circ }\sin {80^ \circ }as (−32)\left( {\dfrac{{-\sqrt 3 }}{2}} \right) using the sum and difference trigonometric formulae.
So, the correct answer is “ (−32)\left( {-\dfrac{{\sqrt 3 }}{2}} \right) ”.

Note : Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides the simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple answers which can be converted into one another. Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths.