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anonymous

  • 5 years ago

If x=4tanθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ<π/2

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  1. anonymous
    • 5 years ago
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    yes, first substitute \[4\tan \theta\] for x

  2. anonymous
    • 5 years ago
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    so you have\[\sqrt{16+16\tan ^{2}\theta}\]

  3. anonymous
    • 5 years ago
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    I am assuming that you would plug in 4tanθ to get: ` √(16+(4tanθ)^2) just not sure where to go from there

  4. anonymous
    • 5 years ago
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    you can factor out a 16 and take it out of the radical \[\sqrt{16(1+\tan ^{2}\theta)} = 4\sqrt{(1+\tan ^{2}\theta)}\]

  5. anonymous
    • 5 years ago
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    now you need to make a trig substitution. \[1+\tan ^{2}\theta=\sec ^{2}\theta\]

  6. anonymous
    • 5 years ago
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    OK... so when you say take it out of the radical you mean...? sorry this is where I get confused

  7. anonymous
    • 5 years ago
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    since we had \[\sqrt{16(1+\tan ^{2}\theta)}\] we can evaluate the square root of 16

  8. anonymous
    • 5 years ago
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    Ohhh ok

  9. anonymous
    • 5 years ago
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    and take it out of the square root

  10. anonymous
    • 5 years ago
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    so then the final step would be \[4\sqrt{\sec ^{2}\theta}=4\sec \theta\]

  11. anonymous
    • 5 years ago
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    does that make sense?

  12. anonymous
    • 5 years ago
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    sort of... I'm trying to figure out how you got 1+tan^2θ=sec^2θ to turn into 4√(sec^2θ)=4secθ?

  13. anonymous
    • 5 years ago
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    ok

  14. anonymous
    • 5 years ago
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    \[4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}\]

  15. anonymous
    • 5 years ago
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    I first just replaced 1+tan^2θ with sec^2θ

  16. anonymous
    • 5 years ago
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    then, we can take the square root of sec^2θ

  17. anonymous
    • 5 years ago
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    which is just secθ

  18. anonymous
    • 5 years ago
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    oh ok that makes sense thanks!

  19. anonymous
    • 5 years ago
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    No problem

  20. anonymous
    • 5 years ago
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    say I had one with cot.... √(9+(3cot^2θ) √(9+9cotθ)?

  21. anonymous
    • 5 years ago
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    or I mean √(9+9cot^2θ)?

  22. anonymous
    • 5 years ago
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    is that right

  23. anonymous
    • 5 years ago
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    ok then I think....1√(1+cot^2θ) how do I know what to substitute then?

  24. anonymous
    • 5 years ago
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    Or I mean 9√(1+cot^2θ)

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