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alexeis_nicole

  • one year ago

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

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  1. bahrom7893
    • one year ago
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    Hint: apply the double angle formula

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

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

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

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

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

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

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

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

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

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

  12. alexeis_nicole
    • one year ago
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    lol i did too

  13. RadEn
    • one year ago
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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

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

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

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

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

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

  19. satellite73
    • one year ago
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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

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