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alexeis_nicole

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

  • one year ago
  • one year ago

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  1. bahrom7893
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    Hint: apply the double angle formula

    • one year ago
  2. alexeis_nicole
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    would i matter which side i start first ? :S

    • one year ago
  3. satellite73
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    i am confused why don't you just add \(\tan(2x)\) to both sides and get rid of it

    • one year ago
  4. bahrom7893
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    because it's times

    • one year ago
  5. satellite73
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    oh because it is times doh

    • one year ago
  6. alexeis_nicole
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    so wait... 2 tanx + 2a - 1 = tan2x - tan2x tan^2x does that work?

    • one year ago
  7. alexeis_nicole
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    if i bring tan2x to the right then bring 1 to the left side?

    • one year ago
  8. satellite73
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    i think it is a bunch of algebra i can try it with pencil and paper if you like

    • one year ago
  9. alexeis_nicole
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    yes please? i want to understand how to do it.. i have a test coming up

    • one year ago
  10. alexeis_nicole
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    im trying to do it myself as well

    • one year ago
  11. satellite73
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    ok i screwed up already, let me try it again

    • one year ago
  12. alexeis_nicole
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    lol i did too

    • one year ago
  13. RadEn
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    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

    • one year ago
  14. satellite73
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    this has nothing whatsoever to do with trig so fora as i can tell

    • one year ago
  15. satellite73
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    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 \[\frac{2x(1-x^2)-2x+2a(1-x^2)}{1-x^2}=\frac{1-x^2-2x^3}{1-x^2}\]

    • one year ago
  16. satellite73
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    ignoring the denominators, you ge t \[2x-2x^3-2x+2a-2ax^2=1-x^2-2x^3\] \[2a-2ax^2=1-x^2\]

    • one year ago
  17. satellite73
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    and therefore \(a=\frac{1}{2}\)

    • one year ago
  18. alexeis_nicole
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    wth! LOL that's the right answer... omg i was so off.... replacing tan(x) with "x" helped. thank you!

    • one year ago
  19. satellite73
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    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

    • one year ago
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