anonymous
  • anonymous
Question on Advanced function grade 12 Determine the value of "a" in the following equation : 2 tanx - tan2x + 2a = 1 - tan2x ・ tan^2x Hint: Double Angle Formula for Tangent is : tan2x = 2tanx/1-tan^2x
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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bahrom7893
  • bahrom7893
Hint: apply the double angle formula
anonymous
  • anonymous
would i matter which side i start first ? :S
anonymous
  • anonymous
i am confused why don't you just add \(\tan(2x)\) to both sides and get rid of it

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bahrom7893
  • bahrom7893
because it's times
anonymous
  • anonymous
oh because it is times doh
anonymous
  • anonymous
so wait... 2 tanx + 2a - 1 = tan2x - tan2x tan^2x does that work?
anonymous
  • anonymous
if i bring tan2x to the right then bring 1 to the left side?
anonymous
  • anonymous
i think it is a bunch of algebra i can try it with pencil and paper if you like
anonymous
  • anonymous
yes please? i want to understand how to do it.. i have a test coming up
anonymous
  • anonymous
im trying to do it myself as well
anonymous
  • anonymous
ok i screwed up already, let me try it again
anonymous
  • anonymous
lol i did too
RadEn
  • RadEn
from : tan2x = 2tanx/(1-tan^2 x) 2tanx = tan2x(1-tan^2 x) 2tanx = tan2x - tan2x tan^2 x now, apply it to left side
anonymous
  • anonymous
this has nothing whatsoever to do with trig so fora as i can tell
anonymous
  • anonymous
if we replace \(\tan(x)\) by \(x\) you get \[2x-\frac{2x}{1-x^2}+2a=1-\frac{2x^3}{1-x^2}\] now we add up on the left and right get \[\frac{2x(1-x^2)-2x+2a(1-x^2)}{1-x^2}=\frac{1-x^2-2x^3}{1-x^2}\]
anonymous
  • anonymous
ignoring the denominators, you ge t \[2x-2x^3-2x+2a-2ax^2=1-x^2-2x^3\] \[2a-2ax^2=1-x^2\]
anonymous
  • anonymous
and therefore \(a=\frac{1}{2}\)
anonymous
  • anonymous
wth! LOL that's the right answer... omg i was so off.... replacing tan(x) with "x" helped. thank you!
anonymous
  • anonymous
yeah that is a good trick to remember when you have a ton of annoying algebra to do it is good for all kinds of trig identities because almost all of the work is algebra, and you get too bogged down writing \(\sin(x)\) and \(\tan(x)\) when \(a\) or \(b\) would do

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