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Question: Find the sum of the expression \(\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\...

Find the sum of the expression 1sin⁡θsin⁡2θ+1sin⁡2θsin⁡3θ+1sin⁡3θsin⁡4θ+...........n\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........n terms

Explanation

Solution

Hint:Multiply and divide the whole expression by sin⁡θ\sin \theta .And split sin⁡θ\sin \theta as sin⁡(2θ−θ)\sin \left( 2\theta -\theta \right) for first term, sin⁡(3θ−2θ)\sin \left( 3\theta -2\theta \right) for second term, sin⁡(4θ−3θ)\sin \left( 4\theta -3\theta \right) for third term and similarly so on. Use the trigonometry identity given as
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B.

Complete step-by-step answer:
Given series in the equation is
1sin⁡θsin⁡2θ+1sin⁡2θsin⁡3θ+1sin⁡3θsin⁡4θ+...........n\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........n
Let sum of the given series be ‘s’, we get
s=1sin⁡θsin⁡2θ+1sin⁡2θsin⁡3θ+1sin⁡3θsin⁡4θ+........... .......(i)s=\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........\text{ }.......\left( i \right)
Multiply and divide by sin⁡θ\sin \theta in the right-hand side of the above equation
So, we get value of s as
s=sin⁡θsin⁡θ[1sin⁡θsin⁡2θ+1sin⁡2θsin⁡3θ+1sin⁡3θsin⁡4θ+...........]s=\dfrac{\sin \theta }{\sin \theta }\left[ \dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+........... \right]
s=1sin⁡θ[sin⁡θsin⁡θsin⁡2θ+sin⁡θsin⁡2θsin⁡3θ+sin⁡θsin⁡3θsin⁡4θ+...........] ......(ii)s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \theta }{\sin \theta \sin 2\theta }+\dfrac{\sin \theta }{\sin 2\theta \sin 3\theta }+\dfrac{\sin \theta }{\sin 3\theta \sin 4\theta }+........... \right]\text{ }......\left( ii \right)
Now, we can write the last term of the given series as
sin⁡θsin⁡(n−1)θsin⁡nθ........(iii)\dfrac{\sin \theta }{\sin \left( n-1 \right)\theta \sin n\theta }........\left( iii \right)
Because, the difference in denominator terms in the product is θ\theta .So, we can write the general term or last term as expression given in the problem
So, we get equation (ii) as
s=1sin⁡θ[sin⁡θsin⁡θsin⁡2θ+sin⁡θsin⁡2θsin⁡3θ+sin⁡θsin⁡3θsin⁡4θ+.........sin⁡θsin⁡(n−1)θsin⁡nθ]......(v)s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \theta }{\sin \theta \sin 2\theta }+\dfrac{\sin \theta }{\sin 2\theta \sin 3\theta }+\dfrac{\sin \theta }{\sin 3\theta \sin 4\theta }+.........\dfrac{\sin \theta }{\sin \left( n-1 \right)\theta \sin n\theta } \right]......\left( v \right)
Now, replace sin⁡θ\sin \theta in numerator of all terms in the bracket as
sin⁡θ=sin⁡(2θ−θ) 1st term sin⁡θ=sin⁡(3θ−2θ) 2nd term sin⁡θ=sin⁡(4θ−3θ) 3rd term \begin{aligned} & \sin \theta =\sin \left( 2\theta -\theta \right)\text{ }{{\text{1}}^{st}}\text{ term} \\\ & \sin \theta =\sin \left( 3\theta -2\theta \right)\text{ }{{\text{2}}^{nd}}\text{ term} \\\ & \sin \theta =\sin \left( 4\theta -3\theta \right)\text{ }{{\text{3}}^{rd}}\text{ term} \\\ \end{aligned}
Similarly, we can write sin⁡θ\sin \theta for nth term as
sin⁡θ=sin⁡(nθ−(n−1)θ) nth term\sin \theta =\sin \left( n\theta -\left( n-1 \right)\theta \right)\text{ }{{\text{n}}^{th}}\text{ term}
Hence, we can write equation (iv) with the help of above equations as
s=1sin⁡θ[sin⁡(2θ−θ)sin⁡θsin⁡2θ+sin⁡(3θ−2θ)sin⁡2θsin⁡3θ+sin⁡(4θ−3θ)sin⁡3θsin⁡4θ+.........sin⁡((nθ)−(n−1)θ)sin⁡(n−1)θsin⁡nθ]s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \left( 2\theta -\theta \right)}{\sin \theta \sin 2\theta }+\dfrac{\sin \left( 3\theta -2\theta \right)}{\sin 2\theta \sin 3\theta }+\dfrac{\sin \left( 4\theta -3\theta \right)}{\sin 3\theta \sin 4\theta }+.........\dfrac{\sin \left( \left( n\theta \right)-\left( n-1 \right)\theta \right)}{\sin \left( n-1 \right)\theta \sin n\theta } \right]
Now use the trigonometric identity given as
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B
Hence, we can write series S using the above identity as

& \dfrac{\sin 2\theta cos\theta -cos2\theta sin\theta }{\sin \theta \sin 2\theta }+\dfrac{\sin 3\theta cos2\theta -cos3\theta sin2\theta }{\sin 2\theta \sin 3\theta } \\\ & \text{ }+\dfrac{\sin 4\theta cos3\theta -cos4\theta sin3\theta }{\sin 3\theta \sin 4\theta }+......\dfrac{\sin \left( n\theta \right)\cos \left( n-1 \right)\theta -\sin \left( n-1 \right)\theta \cos n\theta }{\sin \left( n-1 \right)\theta \sin n\theta } \\\ \end{aligned} \right]$$ $$s=\dfrac{1}{\sin \theta }\left[ \begin{aligned} & \dfrac{\sin 2\theta cos\theta }{\sin \theta \sin 2\theta }-\dfrac{cos2\theta sin\theta }{\sin \theta \sin 2\theta }+\dfrac{\sin 3\theta cos2\theta }{\sin 2\theta \sin 3\theta }-\dfrac{cos3\theta sin2\theta }{\sin 2\theta \sin 3\theta } \\\ & \text{ }+\dfrac{\sin 4\theta cos3\theta }{\sin 3\theta \sin 4\theta }-\dfrac{cos4\theta sin3\theta }{\sin 3\theta \sin 4\theta }+......\dfrac{\sin \left( n\theta \right)\cos \left( n-1 \right)\theta }{\sin \left( n-1 \right)\theta \sin n\theta }-\dfrac{\sin \left( n-1 \right)\theta \cos n\theta }{\sin \left( n-1 \right)\theta \sin n\theta } \\\ \end{aligned} \right]$$ $s=\dfrac{1}{\sin \theta }\left[ \begin{aligned} & \dfrac{cos\theta }{\sin \theta }-\dfrac{cos2\theta }{\sin 2\theta }+\dfrac{cos2\theta }{\sin 2\theta }-\dfrac{cos3\theta }{\sin 3\theta }+\dfrac{cos3\theta }{\sin 3\theta }-\dfrac{cos4\theta }{\sin 4\theta } \\\ & \text{ }+.........\dfrac{\cos \left( n-1 \right)\theta }{\sin \left( n-1 \right)\theta }-\dfrac{\cos n\theta }{\sin n\theta } \\\ \end{aligned} \right]$ We know $\dfrac{\cos x}{\sin x}=\cot x$ So, we get value of ‘s’ as $s=\dfrac{1}{\sin \theta }\left[ \cot \theta -\cot 2\theta +\cot 2\theta -\cot 3\theta +\cot 3\theta -\cot 4\theta +.........\cot \left( n-1 \right)\theta -\operatorname{cotn}\theta \right]$ The above series can also be written as $$s=\dfrac{1}{\sin \theta }\left[ \begin{aligned} & \cot \theta -\cot 2\theta + \\\ & \cot 2\theta -\cot 3\theta + \\\ & \cot 3\theta -\cot 4\theta + \\\ & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\\ & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\\ & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\\ & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\\ & \cot \left( n-1 \right)\theta -\cot n\theta \\\ \end{aligned} \right]$$ So, we get value of ‘s’ as $\begin{aligned} & s=\dfrac{1}{\sin \theta }\left[ \cot \theta -\cot n\theta \right] \\\ & s=\dfrac{1}{\sin \theta }\left[ \dfrac{\cos \theta }{\sin \theta }-\dfrac{\cos n\theta }{\sin n\theta } \right] \\\ & s=\dfrac{1}{\sin \theta }\left[ \dfrac{\cos \theta \sin n\theta -\cos n\theta \sin \theta }{\sin \theta \sin n\theta } \right] \\\ & s=\dfrac{1}{\sin \theta }\dfrac{\sin \left( n\theta -\theta \right)}{\sin \theta \sin n\theta } \\\ & s=\dfrac{\sin \left( n-1 \right)\theta }{{{\sin }^{2}}\theta \sin n\theta } \\\ \end{aligned}$ Hence, sum of given series is $s=\dfrac{\cot \theta -\cot n\theta }{\sin \theta }=\dfrac{\sin \left( n-1 \right)\theta }{{{\sin }^{2}}\theta \sin n\theta }$ Note: If general term of any series is ${{T}_{n}}=\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \sin n\theta }}$ Then divide and multiply the whole equation by $\cos \theta =\cos \left( n\theta -\left( n-1 \right)\theta \right)$ And if ${{T}_{n}}$ is of type $${{T}_{n}}=\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \cos n\theta }\text{ (or) }}\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \sin n\theta }\text{ }}$$ Then divide and multiply the whole equation by $$\sin \left( n\theta -\left( n-1 \right)\theta \right)$$ Calculation is the important side of this question as well. And creating a difference type with the help of observation is the key point of the solution.