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Question: Using the formula: \[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\] Find the value of...

Using the formula: sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
Find the value of sin⁡15∘\sin {15^\circ }
A. 3\sqrt 3
B. 3+1\sqrt 3 + 1
C. 3−122\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
D. 3+122\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}

Explanation

Solution

We have to find the value of the given trigonometric function sin⁡15∘\sin {15^\circ } using the given formula of difference of two angles of a sine function . We solve this question using the knowledge of various values of trigonometric functions for different values of angles . First we will simplify the value of the angle of the given trigonometric function of the difference of two angles . Then replacing the values of the angles in the given formula and putting the exact values for the angles we will get the value for the given trigonometric function.

Complete step by step answer:
Given : To find the value of sin⁡15∘\sin {15^\circ }
Now , we know that the angle can be represented as difference of two angles as :
sin⁡15∘=sin⁡(45∘−30∘)\sin {15^\circ } = \sin \left( {{{45}^\circ } - {{30}^\circ }} \right)
Now , using the given formula for the difference of two angles of sine functions , we get the value of sin⁡15∘\sin {15^\circ } as :
sin⁡15∘=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘\sin {15^\circ } = \sin {45^\circ }\cos {30^\circ } - \cos {45^\circ }\sin {30^\circ }
We also know that , the value of trigonometric functions for different values id given as :
sin⁡45∘=12\sin {45^\circ } = \dfrac{1}{{\sqrt 2 }}
⇒cos⁡45∘=12\Rightarrow \cos {45^\circ } = \dfrac{1}{{\sqrt 2 }}
⇒cos⁡30∘=32\Rightarrow \cos {30^\circ } = \dfrac{{\sqrt 3 }}{2}
⇒sin⁡30∘=12\Rightarrow \sin {30^\circ } = \dfrac{1}{2}
Putting these values in the above formula , we get the value of sin⁡15∘\sin {15^\circ } as :
sin⁡15∘=12×32−12×12\sin {15^\circ } = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}
⇒sin⁡15∘=322−122\Rightarrow \sin {15^\circ } = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}
On solving , we get the value as :
sin⁡15∘=3−122\sin {15^\circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
Thus , the value of sin⁡15∘\sin {15^\circ } is 3−122\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} .

Hence , the correct option is (C)\left( C \right).

Note: We can also find the value of the given trigonometric function by taking different values for the difference of two angles , like we could also have taken the two angles as 60° and 45° . The various trigonometric formulas are given as :
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B
⇒sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
⇒cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\Rightarrow \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B
⇒cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\Rightarrow \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B
All the trigonometric functions are positive in first quadrant , the sin function are positive in second quadrant and rest are negative , the tan function are positive in third quadrant and rest are negative , the cos function are positive in fourth quadrant and rest are negative.