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Question: The value of \[\sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ \] ( A )\[\dfrac{1}{2}...

The value of sin⁡78∘−sin⁡66∘−sin⁡42∘+sin⁡6∘\sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ
( A )12\dfrac{1}{2}
( B ) −12 - \dfrac{1}{2}
( C ) -1
( D ) none of these

Explanation

Solution

Hint : To solve this kind of questions we are going to use the following procedure given
below so that we can make our question solving process as simple as possible and that will be helpful to save our valuable time.
sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x - \sin y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)
After it put the value of sin or cos with angle

** Complete step-by-step answer** :
We are given the following trigonometric expression : sin⁡78∘−sin⁡66∘−sin⁡42∘+sin⁡6∘\sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ

⇒sin⁡78∘−sin⁡66∘−sin⁡42∘+sin⁡6∘=(sin⁡78∘−sin⁡42∘)+(sin⁡6∘−sin⁡66∘)         =(sin⁡(60∘+18∘)−sin⁡(60∘−18∘))+(sin⁡6∘−sin⁡(60∘+6∘))  \Rightarrow \sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ = (\sin 78^\circ - \sin 42^\circ ) + (\sin 6^\circ - \sin 66^\circ ) \\\ \;\;\;\; = (\sin (60^\circ + 18^\circ ) - \sin (60^\circ - 18^\circ )) + (\sin 6^\circ - \sin (60^\circ + 6^\circ )) \;

Apply the above formula

\sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ = 2\cos \left( {\dfrac{{60^\circ + 18^\circ + 60^\circ - 18^\circ }}{2}} \right)\sin \left( {\dfrac{{60^\circ + 18^\circ - 60^\circ + 18^\circ }}{2}} \right) + 2\cos \left( {\dfrac{{6^\circ + 60^\circ + 6^\circ }}{2}} \right)\sin \left( {\dfrac{{6^\circ - 60^\circ - 6^\circ }}{2}} \right) \\\ = 2\cos \left( {60^\circ } \right)\sin \left( {18^\circ } \right) + 2\cos \left( {36^\circ } \right)\sin ( - 30^\circ ) \\\ = 2.\dfrac{1}{2}\sin \left( {18^\circ } \right) - 2.\dfrac{1}{2}\cos \left( {36^\circ } \right) \\\ = \sin \left( {18^\circ } \right) - \cos \left( {36^\circ } \right) \\\ $$Substitute the value of $ \sin \left( {18^\circ } \right) $ and $ \cos \left( {36^\circ } \right) $ in the above equation.

\Rightarrow \sin 78^\circ - \sin 66^\circ - \sin 42^\circ + \sin 6^\circ = \sin \left( {18^\circ } \right) - \cos \left( {36^\circ } \right) \\
= \dfrac{{\sqrt 5 - 1}}{4} - \dfrac{{\sqrt 5 + 1}}{4} \\
= \dfrac{{\sqrt 5 - 1 - \sqrt 5 - 1}}{4} \\
= \dfrac{{ - 2}}{4} \\
= - \dfrac{1}{2} \\

**So, the correct answer is “Option B”.** **Note** : In this type of question we need to take care of the many things and some of them are mentioned here which will be really helpful to understand the concept: We need to use correct formula in such a way that solution does not become too complex Apply the formula $$\sin x - \sin y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)$$ carefully. Value of $ \sin \left( {18^\circ } \right) $ and $ \cos \left( {36^\circ } \right) $ should be correct to get the right answer without any kind of error.