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alexeis_nicole

  • 2 years 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
    • 2 years ago
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    Hint: apply the double angle formula

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

  3. satellite73
    • 2 years 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
    • 2 years ago
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    because it's times

  5. satellite73
    • 2 years ago
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    oh because it is times doh

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

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

  8. satellite73
    • 2 years 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
    • 2 years ago
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    yes please? i want to understand how to do it.. i have a test coming up

  10. alexeis_nicole
    • 2 years ago
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    im trying to do it myself as well

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

  12. alexeis_nicole
    • 2 years ago
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    lol i did too

  13. RadEn
    • 2 years 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
    • 2 years ago
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    this has nothing whatsoever to do with trig so fora as i can tell

  15. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    and therefore \(a=\frac{1}{2}\)

  18. alexeis_nicole
    • 2 years 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
    • 2 years 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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