Question

The value of
sin1 + sin3 + sin5+ ...+ sin89
a)1/sin1 b)2/sin 1 c)1/2sin1 d)1/cos1

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: sin1 + sin3 + sin5 + - - - + sin89$

[ Remark :- Here 1, 3, 5, _____, 89 are in degrees ]

Since, 1, 3, 5, ____, 89 forms an AP series with

First term, a = 1

Common difference, d = 2

nth term = 89

We know,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm \: 89 = 1 + (n - 1)2$

$\rm \: 88 = 2n - 2$

$\rm \: 88 + 2 = 2n$

$\rm \: 90 = 2n$

$\rm\implies \:n = 45$

So,

$\rm \: = \: \underbrace{sin1 + sin3 + sin5 + - - - + sin89} \\ \: \: \: \: 45 \: terms$

We know,

$\rm \: sina + sin(a + d) + sin(a + 2d) + - - + sin(a + (n - 1)d) \\ \rm \: = \: \dfrac{sin\dfrac{nd}{2}sin\bigg( \dfrac{2a + (n - 1)d}{2} \bigg) }{sin\dfrac{d}{2} }$

So, on substituting the values, we get

$\rm \: sin1 + sin3 + sin5 + - - + sin89 \\ \\ \rm \: = \: \dfrac{sin\dfrac{45 \times 2}{2}sin\bigg( \dfrac{2 \times 1 + (45 - 1)\times 2}{2} \bigg) }{sin\dfrac{2}{2} }$

$\rm \: sin1 + sin3 + sin5 + - - + sin89 \\ \\ \rm \: = \: \dfrac{sin45 \times sin45}{sin1}$

$\rm \: sin1 + sin3 + sin5 + - - + sin89 = \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{sin1}$

$\rm \: sin1 + sin3 + sin5 + - - + sin89 = \dfrac{1}{2 \: sin1}$

So, option (c) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\rm \: cosa + cos(a + d) + cos(a + 2d) + - - + cos(a + (n - 1)d) \\ \rm \: = \: \dfrac{sin\dfrac{nd}{2} \: \: cos\bigg( \dfrac{2a + (n - 1)d}{2} \bigg) }{sin\dfrac{d}{2} }$