Question

If we have a trigonometric expression as \[\sin A\sin ({60^ \circ } - A)\sin ({60^ \circ } + A) = k\sin 3A\], then what is k equal to?
1. \[\dfrac{1}{4}\]
2. \[\dfrac{1}{2}\]
3. \[1\]
4. \[4\]

Answer 1

Hint: To find the required value of k we will use the trigonometric identities of \[\sin (A + B) = \sin A\cos B + \cos A\sin B\]and \[\sin (A - B) = \sin A\cos B - \cos A\sin B\] to simplify the given expression. We will put the required basic values of the trigonometric functions and hence after simplifying it using basic arithmetic operations we will get the required answer.

Complete step-by-step solution:
Trigonometric functions are also known as Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions.
The angles of the sine, the cosine, and the tangent are the primary classification of functions of trigonometry. And the three functions which are the cotangent, the secant and the cosecant can be derived from the primary functions.
We are given \[\sin A\sin ({60^ \circ } - A)\sin ({60^ \circ } + A) = k\sin 3A\]
Or we can rewrite it as \[\sin A\left[ {\sin ({{60}^ \circ } - A)\sin ({{60}^ \circ } + A)} \right] = k\sin 3A\]
Using the following identities :
\[\sin (A + B) = \sin A\cos B + \cos A\sin B\]
\[\sin (A - B) = \sin A\cos B - \cos A\sin B\]
We get the following expression
\[\sin A\left[ {\left( {\sin {{60}^ \circ }\cos A - \cos {{60}^ \circ }\sin A} \right)\left( {\sin {{60}^ \circ }\cos A + \cos {{60}^ \circ }\sin A} \right)} \right] = k\sin 3A\]
On simplification we get the following expression
\[\sin A\left( {{{\sin }^2}{{60}^ \circ }{{\cos }^2}A - {{\cos }^2}{{60}^ \circ }{{\sin }^2}A} \right) = k\sin 3A\]
Putting the values of \[\sin {60^ \circ }\] and \[\cos {60^ \circ }\] we get the following expression
\[\sin A\left( {\dfrac{3}{4}{{\cos }^2}A - \dfrac{1}{4}{{\sin }^2}A} \right) = k\sin 3A\]
We know that \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
\[\sin A\left( {\dfrac{3}{4}\left( {1 - {{\sin }^2}A} \right) - \dfrac{1}{4}{{\sin }^2}A} \right) = k\sin 3A\]
On simplification we get the following expression
\[\dfrac{1}{4}\left( {3\sin A - 4{{\sin }^3}A} \right) = k\sin 3A\]
\[\sin A\left( {\dfrac{3}{4} - {{\sin }^2}A} \right) = k\sin 3A\]
Taking \[\dfrac{1}{4}\] common we get the following expression
\[\dfrac{1}{4}\left( {3\sin A - 4{{\sin }^3}A} \right) = k\sin 3A\]
Since \[\sin 3A = 3\sin A - {\sin ^3}A\]
We get \[\dfrac{1}{4}\sin 3A = k\sin 3A\]
Therefore we get \[k = \dfrac{1}{4}\].
Therefore option (1) is the correct answer.

Note: To solve such type of questions one must have a strong grip over the concepts of trigonometry, its related formulas and basic trignometric identities so as to simplify the expression obtained at each step of the calculation. We must do the calculations carefully and should recheck them in order to get the desired result correctly and avoid errors.