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Question: The value of \(\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right)\)...

The value of sin(π10)sin(3π10)\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right) is equal to:
A. 12\dfrac{1}{2}
B. 12 - \dfrac{1}{2}
C. 14\dfrac{1}{4}
D. 11

Explanation

Solution

The given question requires us to find the value of the product of two sine functions with different angles. So, we will do the basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as sin(90x)=cosx\sin \left( {{{90}^ \circ } - x} \right) = \cos x and double angle formula for sine. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the simplification rules to solve the problem with ease.

Complete step by step answer:
In the given problem, we have to find value of sin(π10)sin(3π10)\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right).
So, we have, sin(π10)sin(3π10)\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right)
Using the trigonometric identity sin(90x)=cosx\sin \left( {{{90}^ \circ } - x} \right) = \cos x, we get,
sin(π10)cos(π23π10)\Rightarrow \sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{{10}}} \right)
Simplifying the angle in cosine, we get,
sin(π10)cos(5π3π10)\Rightarrow \sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{{5\pi - 3\pi }}{{10}}} \right)
sin(π10)cos(2π10)\Rightarrow \sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{{2\pi }}{{10}}} \right)

Multiplying and dividing the expression by 2cos(π10)2\cos \left( {\dfrac{\pi }{{10}}} \right), we get,
2cos(π10)2cos(π10)×sin(π10)cos(2π10)\Rightarrow \dfrac{{2\cos \left( {\dfrac{\pi }{{10}}} \right)}}{{2\cos \left( {\dfrac{\pi }{{10}}} \right)}} \times \sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{{2\pi }}{{10}}} \right)
Now, using the double angle formula for sine sin2x=2sinxcosx\sin 2x = 2\sin x\cos x, we get,
sin(2π10)2cos(π10)×cos(2π10)\Rightarrow \dfrac{{\sin \left( {\dfrac{{2\pi }}{{10}}} \right)}}{{2\cos \left( {\dfrac{\pi }{{10}}} \right)}} \times \cos \left( {\dfrac{{2\pi }}{{10}}} \right)
Now, multiplying and dividing the expression by 22, we get,
2sin(2π10)cos(2π10)2×2cos(π10)\Rightarrow \dfrac{{2\sin \left( {\dfrac{{2\pi }}{{10}}} \right)\cos \left( {\dfrac{{2\pi }}{{10}}} \right)}}{{2 \times 2\cos \left( {\dfrac{\pi }{{10}}} \right)}}
Now, using the double angle formula for sine again, we get,
sin(4π10)4cos(π10)\Rightarrow \dfrac{{\sin \left( {\dfrac{{4\pi }}{{10}}} \right)}}{{4\cos \left( {\dfrac{\pi }{{10}}} \right)}}

Now, we know that sine and cosine are complementary trigonometric functions. So, we get,
cos(π24π10)4cos(π10)\Rightarrow \dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{{4\pi }}{{10}}} \right)}}{{4\cos \left( {\dfrac{\pi }{{10}}} \right)}}
Simplifying the expression further, we get,
cos(5π4π10)4cos(π10)\Rightarrow \dfrac{{\cos \left( {\dfrac{{5\pi - 4\pi }}{{10}}} \right)}}{{4\cos \left( {\dfrac{\pi }{{10}}} \right)}}
cos(π10)4cos(π10)\Rightarrow \dfrac{{\cos \left( {\dfrac{\pi }{{10}}} \right)}}{{4\cos \left( {\dfrac{\pi }{{10}}} \right)}}
Cancelling the common factors in numerator and denominator, we get,
14\Rightarrow \dfrac{1}{4}
So, the value of the trigonometric expression sin(π10)sin(3π10)\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right) is 14\dfrac{1}{4}.

Hence, option C is the correct answer.

Note: We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.