Kandarp's Observation

If $ a_1, a_2, a_3, \ldots a_n $ are in an Arithmetic Progression with common difference $ d \neq 0 $, then prove that

$\sin (d) \times ( \csc (a_1) \times \csc (a_2) + \csc (a_2) \times \csc (a_3) + \ldots + \csc (a_{n-1}) \times \csc (a_n) ) = \cot( a_ 1 )- \cot (a_n.)$

#Geometry

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Comments

@Kandarp Singh I believe that you will get better response posting this as a note for discussion, instead of as a question looking for a numerical answer.

Calvin Lin Staff - 6 years, 11 months ago

Kandarp Singh - 6 years, 11 months ago

First open all the terms to get $LHS=sin(d)csc({ a }_{ 1 })csc({ a }_{ 2 })+sin(d)csc({ a }_{ 2 })csc({ a }_{ 3 })+................sin(d)csc({ a }_{ n-1 })csc({ a }_{ n })\\ =\frac { sin(d) }{ sin({ a }_{ 1 })sin({ a }_{ 2 }) } +\frac { sin(d) }{ sin({ a }_{ 2 })sin({ a }_{ 3 }) } +............+\frac { sin(d) }{ sin({ a }_{ n-1 })sin({ a }_{ n }) } .$

Now since ${ a }_{ 1 },{ a }_{ 2 },......{ a }_{ n }$ are in $A.P.$

$d={ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }=............={ a }_{ n }-{ a }_{ n-1 }.$

Now replacing $d$ suitably we get

$LHS=\frac { sin({ a }_{ 2 }-{ a }_{ 1 }) }{ sin({ a }_{ 1 })sin({ a }_{ 2 }) } +\frac { sin({ a }_{ 3 }-{ a }_{ 2 }) }{ sin({ a }_{ 2 })sin({ a }_{ 3 }) } +............+\frac { sin({ a }_{ n }-{ a }_{ n-1 }) }{ sin({ a }_{ n-1 })sin({ a }_{ n }) } .$

Using identity $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$ we get

$\frac { sin(A-B) }{ sin(A)sin(B) } =\frac { sin(A)cos(B)-sin(B)cos(A) }{ sin(A)sin(B) } =cot(B)-cot(A).$

Now our $LHS$ becomes $=\quad (cot({ a }_{ 2 })-cot({ a }_{ 1 }))+(cot({ a }_{ 3 })-cot({ a }_{ 2 }))+...........+(cot({ a }_{ n })-cot({ a }_{ n-1 }))$

Note how terms cancel here hence $LHS$ becomes $=\quad cot({ a }_{ n })-cot({ a }_{ 1 })=RHS\quad$

$Hence\quad proved$

Ronak Agarwal - 6 years, 11 months ago

The $RHS$ kinda suggests that this is telescopical. Indeed, $sin(d)*csc(a_n)*csc(a_{n+1})=\frac {sin(a_{n+1}-a_{n})}{sin(a_n)*sin(a_{n+1})}$ Using the difference formula, it simplifies to $cot(a_n)-cot(a_{n+1})$ .

Hence LHS= $cot(a_1)-cot(a_2)+cot(a_2)-cot(a_3)+...+cot(a_{n-1})-cot{a_n}=cot(a_1)-cot(a_n)$

Xuming Liang - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$