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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.
Ok, so we start with (LHS)
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
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.
Ok, so we start with (LHS)
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
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]
Original post by Sajee
cot 2x + cosec 2x = cot x
how would one go about proving this?
thanks
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.
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.
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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