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Question: If we have a trigonometric expression as \[\sin A\sin ({60^ \circ } - A)\sin ({60^ \circ } + A) = k\...

If we have a trigonometric expression as sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=ksin⁡3A\sin A\sin ({60^ \circ } - A)\sin ({60^ \circ } + A) = k\sin 3A, then what is k equal to?
1. 14\dfrac{1}{4}
2. 12\dfrac{1}{2}
3. 11
4. 44

Explanation

Solution

To find the required value of k we will use the trigonometric identities of sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin Band sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\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⁡Asin⁡(60∘−A)sin⁡(60∘+A)=ksin⁡3A\sin A\sin ({60^ \circ } - A)\sin ({60^ \circ } + A) = k\sin 3A
Or we can rewrite it as sin⁡A[sin⁡(60∘−A)sin⁡(60∘+A)]=ksin⁡3A\sin A\left[ {\sin ({{60}^ \circ } - A)\sin ({{60}^ \circ } + A)} \right] = k\sin 3A
Using the following identities :
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B
We get the following expression
sin⁡A[(sin⁡60∘cos⁡A−cos⁡60∘sin⁡A)(sin⁡60∘cos⁡A+cos⁡60∘sin⁡A)]=ksin⁡3A\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(sin⁡260∘cos⁡2A−cos⁡260∘sin⁡2A)=ksin⁡3A\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∘\sin {60^ \circ } and cos⁡60∘\cos {60^ \circ } we get the following expression
sin⁡A(34cos⁡2A−14sin⁡2A)=ksin⁡3A\sin A\left( {\dfrac{3}{4}{{\cos }^2}A - \dfrac{1}{4}{{\sin }^2}A} \right) = k\sin 3A
We know that cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
sin⁡A(34(1−sin⁡2A)−14sin⁡2A)=ksin⁡3A\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
14(3sin⁡A−4sin⁡3A)=ksin⁡3A\dfrac{1}{4}\left( {3\sin A - 4{{\sin }^3}A} \right) = k\sin 3A
sin⁡A(34−sin⁡2A)=ksin⁡3A\sin A\left( {\dfrac{3}{4} - {{\sin }^2}A} \right) = k\sin 3A
Taking 14\dfrac{1}{4} common we get the following expression
14(3sin⁡A−4sin⁡3A)=ksin⁡3A\dfrac{1}{4}\left( {3\sin A - 4{{\sin }^3}A} \right) = k\sin 3A
Since sin⁡3A=3sin⁡A−sin⁡3A\sin 3A = 3\sin A - {\sin ^3}A
We get 14sin⁡3A=ksin⁡3A\dfrac{1}{4}\sin 3A = k\sin 3A
Therefore we get k=14k = \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.