Question

Prove that the average of the numbers \[n\sin {n^ \circ },n = 2,4,6,...,180\] is \[\cot {1^ \circ }\].

Answer 1

Hint:
First, we will determine the total number of terms. Then, we will put the value of n in the given equation. Then, we will apply trigonometric identities. On simplification, we will get the result.

Complete step by step solution:
Let n be the total number of terms, a be the first term, \[{a_n}\] be the last term and d is the difference between two terms.
Therefore,
 \[
  a = 2 \\
  {a_n} = 180 \\
  d = 2 \\
\]
Total number of terms
 \[ \Rightarrow n = \dfrac{{{a_n} - a}}{d} + 1\]
On substituting the values, we get
 \[ \Rightarrow n = \dfrac{{180 - 2}}{2} + 1\]
On simplification we get
 \[ \Rightarrow n = \dfrac{{178}}{2} + 1\]
On division we get
 \[ \Rightarrow n = 89 + 1\]
Hence, we have
 \[ \Rightarrow n = 90\]
Hence, the total number of terms are 90.
Given,
 \[n\sin {n^ \circ };n = 2,4,6,...,180\]
Therefore, \[n\sin {n^ \circ } = 2\sin {2^ \circ } + 4\sin {4^ \circ } + 6\sin {6^ \circ } + ... + 180\sin {180^ \circ }\] …. (1)
We know that,
 \[\sin \theta = \sin ({180^ \circ } - \theta )\]
And \[\sin {180^ \circ } = 0\]
 \[ \Rightarrow \sin {178^ \circ } = \sin {(180 - 2)^ \circ }\]
So, we have
 \[ \Rightarrow \sin {178^ \circ } = \sin {2^ \circ }\]
Similarly,
 \[ \Rightarrow \sin {176^ \circ } = \sin {(180 - 4)^ \circ }\]
We have,
 \[ \Rightarrow \sin {176^ \circ } = \sin {4^ \circ }\]
Similarly, we have
 \[ \Rightarrow \sin {94^ \circ } = \sin {86^ \circ }\]
 \[ \Rightarrow \sin {92^ \circ } = \sin {88^ \circ }\]
Putting all these values in equation (1)
 \[ \Rightarrow n\sin {n^ \circ } = 2\sin {2^ \circ } + 4\sin {4^ \circ } + 6\sin {6^ \circ } + ... + 180\sin {180^ \circ }\]
We can write the equation as, \[ \Rightarrow n\sin {n^ \circ } = 2\sin {2^ \circ } + 4\sin {4^ \circ } + 6\sin {6^ \circ } + ... + 88\sin {88^ \circ } + 90\sin {90^ \circ } + 92\sin {92^ \circ } + 94\sin {94^ \circ } + ....178\sin {178^ \circ } + 180\sin {180^ \circ }\]
On substituting the values, we get
 \[ = 2\sin {2^ \circ } + 4\sin {4^ \circ } + 6\sin {6^ \circ } + .... + 88\sin {88^ \circ } + 90 \times 1 + 92\sin {88^ \circ } + 94\sin {86^ \circ } + .... + 178\sin {2^ \circ } + 180 \times 0\]
On taking like terms common we get
 \[ = 2\sin {2^ \circ } + 178\sin {2^ \circ } + 4\sin {4^ \circ } + 176\sin {4^ \circ } + 6\sin {6^ \circ } + 174\sin {6^ \circ } + ... + 88\sin {88^ \circ } + 92\sin {88^ \circ } + 90\]
On adding the like terms, we get
 \[ = (2 + 178)\sin {2^ \circ } + (4 + 176)\sin {4^ \circ } + (6 + 174)\sin {6^ \circ } + ... + (88 + 92)\sin {88^ \circ } + 90\]
So, we have
 \[ = 180\sin {2^ \circ } + 180\sin {4^ \circ } + 180\sin {6^ \circ } + ... + 180\sin {88^ \circ } + 90\]
Taking 180 common we get,
 \[ = 180(\sin {2^ \circ } + \sin {4^ \circ } + \sin {6^ \circ } + ... + \sin {88^ \circ }) + 90\]
Now Taking 90 common we get
 \[ = 90(2\sin {2^ \circ } + 2\sin {4^ \circ } + 2\sin {6^ \circ } + ... + 2\sin {88^ \circ } + 1)\]
Average of 90 terms is given as
 \[ \Rightarrow Average = \dfrac{{90(2\sin {2^ \circ } + 2\sin {4^ \circ } + 2\sin {6^ \circ } + ... + 2\sin {{88}^ \circ } + 1)}}{{90}}\]
So, we have
 \[ \Rightarrow Average = 2\sin {2^ \circ } + 2\sin {4^ \circ } + 2\sin {6^ \circ } + ... + 2\sin {88^ \circ } + 1\] …. (2)
Using the formula,
 \[\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)\]
Taking C as \[n - 1\] and D as \[n + 1\],
 \[ \Rightarrow \cos {(n - 1)^ \circ } - \cos {(n + 1)^ \circ } = 2\sin {\left( {\dfrac{{n - 1 + n + 1}}{2}} \right)^ \circ }\sin {\left( {\dfrac{{n - 1 - n - 1}}{2}} \right)^ \circ }\]
On simplification we get,
 \[ \Rightarrow \cos {(n - 1)^ \circ } - \cos {(n + 1)^ \circ } = 2\sin {n^ \circ }\sin {1^ \circ }\]
On dividing by sin1\[^ \circ \] we get,
 \[ \Rightarrow 2\sin {n^ \circ } = \dfrac{{\cos {{(n - 1)}^ \circ } - \cos {{(n + 1)}^ \circ }}}{{\sin {1^ \circ }}}\] …. (3)
Using equation (3) in equation (2)
 \[ \Rightarrow 2\sin {2^ \circ } + 2\sin {4^ \circ } + 2\sin {6^ \circ } + ... + 2\sin {88^ \circ } + 1\]
We have,
 \[ = \dfrac{{\cos {1^ \circ } - \cos {3^ \circ }}}{{\sin {1^ \circ }}} + \dfrac{{\cos {3^ \circ } - \cos {5^ \circ }}}{{\sin {1^ \circ }}} + \dfrac{{\cos {5^ \circ } - \cos {7^ \circ }}}{{\sin {1^ \circ }}} + ... + \dfrac{{\cos {{87}^ \circ } - \cos {{89}^ \circ }}}{{\sin {1^ \circ }}} + 1\]
Taking \[\sin {1^ \circ }\] as L.C.M, we get
 \[ = \dfrac{{\cos {1^ \circ } - \cos {3^ \circ } + \cos {3^ \circ } - \cos {5^ \circ } + \cos {5^ \circ } - \cos {7^ \circ } + ... + \cos {{87}^ \circ } - \cos {{89}^ \circ }}}{{\sin {1^ \circ }}} + 1\]
On simplification,
 \[ = \dfrac{{\cos {1^ \circ } - \cos {{89}^ \circ }}}{{\sin {1^ \circ }}} + 1\]
Using, \[\cos \theta = \sin ({90^ \circ } - \theta )\]
 We get, \[\cos {89^ \circ } = \sin {1^ \circ }\]
 \[ = \dfrac{{\cos {1^ \circ } - \sin {1^ \circ }}}{{\sin {1^ \circ }}} + 1\]
On simplification, we get
 \[ = \cot {1^ \circ } - 1 + 1\]
Hence, we have
 \[ = \cot {1^ \circ }\]

Hence, the average of the numbers \[n\sin {n^ \circ },n = 2,4,6,...,180\] is \[\cot {1^ \circ }\].

Note:
You can get trouble in converting the angles. So, here are some formulas for conversion
1) \[\sin (180 - \theta ) = \sin \theta \]
2) \[\cos (180 - \theta ) = - \cos \theta \]
3) \[\sin \theta = \cos (90 - \theta )\]
4) \[\cos \theta = \sin (90 - \theta )\]