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klgyal
Prove the identity :
2cot2x+tanx=cot x
2cot2x+tanx=cot x
Start at the LHS, and change 2cot2x to 2/tan2x.
Find a common denominator then expand and simplify.
Recommend starting by using identity
If this is subbed in you get
which is then solvable...
Tip - multiply through by tan(X)
If this is subbed in you get
which is then solvable...
Tip - multiply through by tan(X)
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
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
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
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
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
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
Original post by funkmaster39
Identities Proof as i said would produce =]
Absolute legend mate 👍
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