anonymous
  • anonymous
If x=4tanθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ<π/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
yes, first substitute \[4\tan \theta\] for x
anonymous
  • anonymous
so you have\[\sqrt{16+16\tan ^{2}\theta}\]
anonymous
  • anonymous
I am assuming that you would plug in 4tanθ to get: ` √(16+(4tanθ)^2) just not sure where to go from there

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More answers

anonymous
  • anonymous
you can factor out a 16 and take it out of the radical \[\sqrt{16(1+\tan ^{2}\theta)} = 4\sqrt{(1+\tan ^{2}\theta)}\]
anonymous
  • anonymous
now you need to make a trig substitution. \[1+\tan ^{2}\theta=\sec ^{2}\theta\]
anonymous
  • anonymous
OK... so when you say take it out of the radical you mean...? sorry this is where I get confused
anonymous
  • anonymous
since we had \[\sqrt{16(1+\tan ^{2}\theta)}\] we can evaluate the square root of 16
anonymous
  • anonymous
Ohhh ok
anonymous
  • anonymous
and take it out of the square root
anonymous
  • anonymous
so then the final step would be \[4\sqrt{\sec ^{2}\theta}=4\sec \theta\]
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
sort of... I'm trying to figure out how you got 1+tan^2θ=sec^2θ to turn into 4√(sec^2θ)=4secθ?
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}\]
anonymous
  • anonymous
I first just replaced 1+tan^2θ with sec^2θ
anonymous
  • anonymous
then, we can take the square root of sec^2θ
anonymous
  • anonymous
which is just secθ
anonymous
  • anonymous
oh ok that makes sense thanks!
anonymous
  • anonymous
No problem
anonymous
  • anonymous
say I had one with cot.... √(9+(3cot^2θ) √(9+9cotθ)?
anonymous
  • anonymous
or I mean √(9+9cot^2θ)?
anonymous
  • anonymous
is that right
anonymous
  • anonymous
ok then I think....1√(1+cot^2θ) how do I know what to substitute then?
anonymous
  • anonymous
Or I mean 9√(1+cot^2θ)

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