Question

Help needed. Intriguing Question this one. Q. attached.

Answer 1

@amistre64 @ganeshie8 @KingGeorge

Answer 2

Try to change the given angles to Sub-multiple in the first step

Answer 3

1/sin 9 ?
1/(2sin(4.5).cos(4.5)) ?

Answer 4

Try doing that

Answer 5

Okay. Then what?

Answer 6

convert cos 9 also

Answer 7

Which formula should i use?
There are like 3 submultiple formulae for 'cosine'.

Answer 8

@AkashdeepDeb I am unable to clarify. Please ask some one else. Sorry Yaar

Answer 9

It's okay. :)
Thanks for trying.

@Hero ?

Answer 10

To be honest, I don't see a quick and easy way to prove this.

Here's what I've tried (or thought of) so far:

1) \[\frac{ 1 }{ \sin 9 }+\frac{ 1 }{ \cos 9 }=\frac{ \sin9 +\cos 9}{ \sin9\cos 9 }=\sqrt{m}+\sqrt{n}\]

2) Because sin 2x = 2 sin x cos x, 2 sin 9 cos 9 = sin 18. Therefore, \[\sin9\cos9=\sin (2*9)=\frac{ \sin18 }{ 2 }\]

Answer 11

This is not necessarily the right thing to do, but it may help you get started.

I find myself wondering whether there's an alternative way to write sin 9 + cos 9.

Keep in mind that our end goal is to find the values of m and n, which must be positive integers.

Answer 12

Okay, I guess then:
\[\frac{2.sin 9}{sin 18} + \frac{2cos9}{sin 18}\]

Now we cannot say that \[\sqrt{m} = \frac{2sin9}{sin 18} ~~~and~~~\sqrt{n} = \frac{2cos9}{cos18}\] right? :P

Answer 13

If this is supposed to be an approximation, I think that should work out with m=1,n=4.
But if it is supposed to be exact. It doesn't.

Answer 14

Ooops..I meant to right sin 18 instead of cos 18 over there. :-)

Answer 15

Okay no, then \(n\neq4\)

Answer 16

My main suggestion at this point, Akashdeep, is that you find / obtain a table of trig identities and look for one or more that involves the Sqrt operator, as well as possibily half-angle or double-angles.

Answer 17

\[\sqrt{m} = \frac{2\sin9}{\sin 18} ~~~and~~~\sqrt{n} = \frac{2\cos9}{\sin18}\] Here I've fixed your typo.

Answer 18

Okay. I guess I'll try that then.
Thanks for helping! :)

Answer 19

Wish I could be of more help, but you sound very capable and I have faith in your ability to find a solution here.

consider the slightly more general form\[\sqrt{m} = \frac{2sina}{\sin 2a} ~~~and~~~\sqrt{n} = \frac{2cosa}{\cos2a}\] as you look for identities that might apply here. Good luck, Akashdeep!

Answer 20

@amistre64
@jim_thompson5910

Answer 21

@phi @AravindG

Answer 22

It's certainly worth trying alternative approaches, such as De Moivre's theorem or Euler's formula. In any case, spending some time doing that might very well turn out to be faster than waiting for help here on OpenStudy.

I look forward to working with you again.

Answer 23

\[

\dfrac{1}{\sin 9} + \dfrac{1}{\cos 9} = \dfrac{2}{\sin 18} \left(\cos 9 + \sin 9\right) = \dfrac{2\sqrt{2}}{\sin 18} \left(\cos 9 \cos 45 + \sin 9 \sin 45\right) \\ =\dfrac{2\sqrt{2}}{\sin 18} \cos36 = \dfrac{2\sqrt{2}}{\sin 18} \sin(54) \\ =2\sqrt{2}\left(3-4\sin^218\right) = \cdots = \sqrt{18} + \sqrt{10}


\]

Answer 24

What are the steps in the "\(...\)" ?

Answer 25

that's for you to fill :)
https://in.answers.yahoo.com/question/index?qid=20061025091609AASJNUn

Answer 26

I think multiplying by 2 and dividing by 2 helps