Question

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

Answer 1

yes, first substitute \[4\tan \theta\] for x

Answer 2

so you have\[\sqrt{16+16\tan ^{2}\theta}\]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

OK... so when you say take it out of the radical you mean...? sorry this is where I get confused

Answer 7

since we had \[\sqrt{16(1+\tan ^{2}\theta)}\] we can evaluate the square root of 16

Answer 8

Ohhh ok

Answer 9

and take it out of the square root

Answer 10

so then the final step would be
\[4\sqrt{\sec ^{2}\theta}=4\sec \theta\]

Answer 11

does that make sense?

Answer 12

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

Answer 13

ok

Answer 14

\[4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}\]

Answer 15

I first just replaced 1+tan^2θ with sec^2θ

Answer 16

then, we can take the square root of sec^2θ

Answer 17

which is just secθ

Answer 18

oh ok that makes sense thanks!

Answer 19

No problem

Answer 20

say I had one with cot....
√(9+(3cot^2θ)
√(9+9cotθ)?

Answer 21

or I mean √(9+9cot^2θ)?

Answer 22

is that right

Answer 23

ok then I think....1√(1+cot^2θ)
how do I know what to substitute then?

Answer 24

Or I mean 9√(1+cot^2θ)