Question

Use a trigonometric substitution involving tangent to write √(961 + x^2)

as a function of theta in the first quadrant without a radical
So confused pleaseeee help

Answer 1

http://www.wolframalpha.com/input/?i=%E2%88%9A%28961+%2B+x%5E2%29

Answer 2

would it help to know that \(\sqrt{961}=31\)?

Answer 3

put \(x=32\tan(\theta)\) so \(x^2=961\tan^2(\theta)\) and you get
\[\sqrt{961+x^2}=\sqrt{961+961\tan^2(\theta)}=\sqrt{961(1+\tan^2(\theta)}=31\sqrt{1+\tan^2(\theta)}\]

Answer 4

sorry i meant
\[x=31\tan(\theta)\]

Answer 5

and now a miracle occurs, namely that
\[1+\tan^2(\theta)=\sec^2(\theta)\] so you can get rid of the radical

Answer 6

when you see
\[\sqrt{a^2+x^2}\] use
\[x=a\tan(\theta)\] and when you see
\[\sqrt{a^2-x^2}\] use
\[x=a\sin(\theta)\]

Answer 7

we can do another if you like