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Question: Find the value of \(\sin {36^0}\sin {72^0}\sin {108^0}\sin {144^0}\)...

Find the value of sin⁡360sin⁡720sin⁡1080sin⁡1440\sin {36^0}\sin {72^0}\sin {108^0}\sin {144^0}

Explanation

Solution

This type of trigonometric problem will be solved by breaking the functions into different combinations such that it matches the formula. In this question, we will use the formula which is sin⁡(1800−θ)=sin⁡θ\sin \left( {{{180}^0} - \theta } \right) = \sin \theta . So with the help of this, we will solve this problem.

Formula used:
sin⁡(1800−θ)=sin⁡θ\sin \left( {{{180}^0} - \theta } \right) = \sin \theta
And the other formula which we will use is
2sin⁡Asin⁡B=[cos⁡(A−B)−cos⁡(A+B)]2\sin A\sin B = \left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]

Complete step-by-step answer:
Since we have
sin⁡360sin⁡720sin⁡1080sin⁡1440\sin {36^0}\sin {72^0}\sin {108^0}\sin {144^0}
As we know, sin⁡(1800−θ)=sin⁡θ\sin \left( {{{180}^0} - \theta } \right) = \sin \theta
So by using the above formula we can write the equation as
⇒sin⁡360sin⁡720sin⁡(1800−720)sin⁡(1800−360)\Rightarrow \sin {36^0}\sin {72^0}\sin \left( {{{180}^0} - {{72}^0}} \right)\sin \left( {{{180}^0} - {{36}^0}} \right)
So on simplifying the above equation, it can be written as
⇒sin⁡360sin⁡720sin⁡720sin⁡360\Rightarrow \sin {36^0}\sin {72^0}\sin {72^0}\sin {36^0}
And it can also be written as
⇒[sin⁡360sin⁡720]2\Rightarrow {\left[ {\sin {{36}^0}\sin {{72}^0}} \right]^2}
Now we will reduce the above equation in such a way that it will justify the trigonometric formulas
⇒14[2sin⁡360sin⁡720]2\Rightarrow \dfrac{1}{4}{\left[ {2\sin {{36}^0}\sin {{72}^0}} \right]^2}
And as we know, 2sin⁡Asin⁡B=[cos⁡(A−B)−cos⁡(A+B)]2\sin A\sin B = \left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]
On equating with the above formula, we get
⇒14[cos⁡360−cos⁡1080]2\Rightarrow \dfrac{1}{4}{\left[ {\cos {{36}^0} - \cos {{108}^0}} \right]^2}
Since, cos⁡1080=cos⁡(900+180)=−sin⁡180\cos {108^0} = \cos \left( {{{90}^0} + {{18}^0}} \right) = - \sin {18^0}
So putting this in the above equation, we get
⇒14[cos⁡360+sin⁡180]2\Rightarrow \dfrac{1}{4}{\left[ {\cos {{36}^0} + \sin {{18}^0}} \right]^2}
Now putting the valuecos⁡360=(5+1)/4\cos {36^0} = \left( {\sqrt 5 + 1} \right)/4, and sin⁡180=(5−1)/4\sin {18^0} = \left( {\sqrt 5 - 1} \right)/4
We get
⇒14[(5+1)/4+(5−1)/4]2\Rightarrow \dfrac{1}{4}{\left[ {\left( {\sqrt 5 + 1} \right)/4 + \left( {\sqrt 5 - 1} \right)/4} \right]^2}
Now on solving the above equation, we get
⇒14[5/2]2\Rightarrow \dfrac{1}{4}{\left[ {\sqrt 5 /2} \right]^2}
And again further solving more, we get
⇒14[5/4]\Rightarrow \dfrac{1}{4}\left[ {5/4} \right]
Now on the multiplication of it, we get
⇒516\Rightarrow \dfrac{5}{{16}}
Therefore, 516\dfrac{5}{{16}} will be the correct answer.

Note: So to solve the trigonometric problem easily first of all we should learn all the trigonometric functions by heart as well as identities. Then we have to just understand how it came just to understand derivation. Second, we should know every possible conversion of one function into another. Solve as many questions as possible which is related to trigonometry. Also, we should try to solve every question multiple times using different formulas. We should also solve questions using a difficult method rather than an easy method. With the help of all these, we can easily solve trigonometric problems.