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# Trignometric identity watch

1. Can someone please prove this identity:

cot x − cot 2x ≡ cosec 2x

2. The key is to remember the identity,
sin 2x = 2 cos x sin x
and thus,

sin x = sin 2x / 2 cos x (1)

and

cos 2x = 2cos^2 x - 1 (2)

and remember to work in cosines and sines.

cot x - cot 2x

(cos x/sin x) - (cos 2x/sin 2x)

(cos x/(sin 2x / 2 cos x)) - (cos 2x/sin 2x) (using 1)

(2 cos^2/sin 2x) - (cos 2x/sin 2x)

(2 cos^2/sin 2x) - (2cos^2 x - 1/sin 2x) (using 2)

1 / sin 2x

Which is cosec 2x. LHS=RHS. Just a few lines
3. (Original post by aiman)
cot x − cot 2x ≡ cosec 2x
cot x − cot 2x ≡ cosec 2x
1/tanx − 1/tan[2x]
=> cosx/sinx − cos[2x]/sin[2x]
= (cosxsin[2x] − cos[2x]sinx) / sin[2x]sinx
= (2cos²[x]sin[x] − (1-2sin²x)sinx) / sin[2x]sinx
= (2cos²[x]sin[x] − sinx + 2sin³x)) / sin[2x]sinx
= (2(1-sin²[x])sin[x] − sinx + 2sin³x)) / sin[2x]sinx
= (2sinx - 2sin³[x] − sinx + 2sin³x) / sin[2x]sinx
= (2 - 2sin²[x] − 1 + 2sin²x) / sin[2x]
=> 1 / sin[2x] => cosec[2x]

∴ cot x − cot 2x ≡ cosec[2x]
4. Mine was better :P lol
5. Thanks alot both of you were wonderful!
6. cot 2x + cosec 2x = cot x

how would one go about proving this?

thanks
7. (Original post by Sajee)
cot 2x + cosec 2x = cot x

how would one go about proving this?

thanks
This is exactly the same as the problem in the first post of this (six year old) thread, just rearranged; a complete worked answer is provided a few posts up.
8. (Original post by nuodai)
This is exactly the same as the problem in the first post of this (six year old) thread, just rearranged; a complete worked answer is provided a few posts up.
Slightly worrying that the thread dredger didn't notice this.
9. (Original post by nuodai)
This is exactly the same as the problem in the first post of this (six year old) thread, just rearranged; a complete worked answer is provided a few posts up.
thank you,

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Updated: December 6, 2011
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