Question

If $ \sin (C + D) = \sin C \cdot \operatorname{Sin} D + \cos C \cdot \cos D $ then the value of $ \sin {75^ \circ } $ is
(a) $ \dfrac{1}{{2\sqrt 2 }}(\sqrt {3 + 1} ) $
(b) $ \dfrac{1}{2}(\sqrt {3 - 1} ) $
(c) $ \dfrac{{\sqrt 3 }}{2} $
(d) $ \dfrac{1}{2} $

Answer 1

Hint: As we know that the above question is of trigonometry as sine and cosine are trigonometric ratios. Here we have to find the value of $ \sin 75 $ and since it is not a standard angle, we will convert it into the summation of two angles and then we will solve it on the basis of the given trigonometric formula. It means that we will convert it into the sine compound angles. We know that an angle is compound angle when that angle is made by the addition or subtraction of two angles.

Complete step-by-step answer:
As per the question we have been given
 $ \sin (C + D) = \sin C \cdot \operatorname{Sin} D + \cos C \cdot \cos D $
and we have to find the value of $ \sin 75 $ .
Now we will break the angle of sine into a sum of two standard angles. We can write
 $\Rightarrow \sin {75^ \circ } = \sin ({45^ \circ } + {30^ \circ }) $ .
By using the property of sum of sine angles we get that:
 $\Rightarrow \sin (45 + 30) = \sin 45\cos 30 + \cos 45\sin 30 $
Now we know the values:
 $\Rightarrow \sin {45^ \circ } = \cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sin {30^ \circ } = \dfrac{1}{2} $ and $ \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2} $ .
Now we substitute these values in the above equation and we have:
 $\Rightarrow \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} $ .
Hence the correct option is (a) $ \dfrac{1}{{2\sqrt 2 }}(\sqrt {3 + 1} ) $ .
So, the correct answer is “Option a”.

Note: We should note that whenever we get such types of questions we should first break the given angle into the sum of two standard angles, and then we apply the basic trigonometric property which is also given in the question this time. We should always remember the basic standard value of angles because we need to apply them always to get the required value of a given angle. And then we solve it.