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Question: The value of \(\sin \left( {60^\circ + \theta } \right) - \cos \left( {30^\circ - \theta } \right)\)...

The value of sin⁡(60∘+θ)−cos⁡(30∘−θ)\sin \left( {60^\circ + \theta } \right) - \cos \left( {30^\circ - \theta } \right) is equal to
A.2cos⁡θ2\cos \theta
B.2sin⁡θ2\sin \theta
C.0
D.1

Explanation

Solution

Hint- In this particular type of question we need to use the trigonometric formula for sin(a+b) and cos(a-b). To further solve it we need to expand and simplify to get the desired answer.

Complete step-by-step answer:
Since
sin⁡(a+b)=sin⁡a.cos⁡b+cos⁡a.sin⁡b and cos(a−b)=cos⁡a.cos⁡b+sin⁡a.sin⁡b  \sin \left( {a + b} \right) = \sin a.\cos b + \cos a.\sin b \\\ and{\text{ cos}}\left( {a - b} \right) = \cos a.\cos b + \sin a.\sin b \\\

sin⁡(60∘+θ)−cos⁡(30∘−θ) =sin⁡60∘.cos⁡θ+cos⁡60∘.sinθ−(cos⁡30∘.cos⁡θ+sin⁡30∘.sin⁡θ) =32.cos⁡θ+12.sin⁡θ−32.cos⁡θ−12sin⁡θ =0  \sin \left( {60^\circ + \theta } \right) - \cos \left( {30^\circ - \theta } \right) \\\ = \sin 60^\circ .\cos \theta + \cos 60^\circ .sin\theta - \left( {\cos 30^\circ .\cos \theta + \sin 30^\circ .\sin \theta } \right) \\\ = \dfrac{{\sqrt 3 }}{2}.\cos \theta + \dfrac{1}{2}.\sin \theta - \dfrac{{\sqrt 3 }}{2}.\cos \theta - \dfrac{1}{2}\sin \theta \\\ = 0 \\\

Note-
Remember to recall the basic trigonometric formulas while solving such types of questions. Note that this question could also be solved by directly converting sin⁡(60∘+θ)\sin \left( {60^\circ + \theta } \right) into sin⁡(90∘−(30∘−θ))\sin \left( {90^\circ - \left( {30^\circ - \theta } \right)} \right) and further into cos⁡(30∘−θ)\cos \left( {30^\circ - \theta } \right) , thus cancelling out both the terms to get 0.