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# C3 Trig watch

1. Prove the identity :

2cot2x+tanx=cot x
2. Take the LHS; can you write 2cot(2x) in terms of tan(x)?
3. (Original post by klgyal)
Prove the identity :

2cot2x+tanx=cot x
Start at the LHS, and change 2cot2x to 2/tan2x.
Find a common denominator then expand and simplify.
4. i got (2+tan2xtanx)/tan2x ?? whats next please ??
5. Recommend starting by using identity

If this is subbed in you get

which is then solvable...

Tip - multiply through by tan(X)
6. (Original post by klgyal)
i got (2+tan2xtanx)/tan2x ?? whats next please ??
I'm not sure what you've done...

2/tan2x+tanx

(2/2tanx/1-tan^x) +tanx

(2(1-tan^2x)/2tanx) +tanx

2-2tan^2x +tan^2x/2tanx

2/2tanx->1/tanx->cotx
7. I actually hate the idea of having to first code trig into notepad ascii, then try and have someone who is desperately looking for an answer for them (who is stressed enough they dont get it) having to try and decode that line after line and see why it works... Tell you what, ill do a working out in a bit, scan it and post it- providing thats possible? :P
8. Identities Proof as i said would produce =]
Attached Images

9. Oh I was just doing this question! Exciting. Well, a couple of hours ago.
10. LHS:

2cot2x + tanx
= (2cosxcos2x + sinxsin2x)/(sin2xcosx)
= [ 2cosx(cos^2x - sin^2x) + sinx2sinxcosx ]/[2sinxcosxcosx]

Expand the brackets

= (2cos^3x - 2cosxsin^2x + 2sin^2xcosx)/(2sinxcosxcosx)

Take out cosx from top and bottom for each term

= (cos^2x - sin^2x + sin^2x)/(sinxcosx)

Tidy it up

= (cos^2x)/(sinxcosx)

Cancel the cosx

= (cosx)/(sinx)

= cotx
11. (Original post by Chloeee02)
Oh I was just doing this question! Exciting. Well, a couple of hours ago.

Same.

12. (Original post by *MJ*)

Same.

Must be everyone's mad revision for C3. Hooray.
13. (Original post by Chloeee02)
Must be everyone's mad revision for C3. Hooray.
Was yours an OCR Soloman paper?

14. (Original post by *MJ*)
Was yours an OCR Soloman paper?

Edexcel Soloman. Guess they're pretty similar though.
15. (Original post by Chloeee02)
Edexcel Soloman. Guess they're pretty similar though.
Nice.

Good luck.
16. (Original post by *MJ*)
Nice.

Good luck.
And youu .
I just need to buy another scientific calculator before Wednesday now...I keep on forgetting to.
17. (Original post by 4WhenImBanned)
LHS:

2cot2x + tanx
= (2cosxcos2x + sinxsin2x)/(sin2xcosx)
= [ 2cosx(cos^2x - sin^2x) + sinx2sinxcosx ]/[2sinxcosxcosx]

Expand the brackets

= (2cos^3x - 2cosxsin^2x + 2sin^2xcosx)/(2sinxcosxcosx)

Take out cosx from top and bottom for each term

= (cos^2x - sin^2x + sin^2x)/(sinxcosx)

Tidy it up

= (cos^2x)/(sinxcosx)

Cancel the cosx

= (cosx)/(sinx)

= cotx
thanks

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