# Need help simplifying this trigonometric substitution problem on homework.

• Apr 10th 2013, 12:46 PM
AKPunk49
Need help simplifying this trigonometric substitution problem on homework.
I need help simplifying (sqrt(x^2-49))/(x) where x =7sec(theta)

I'm completely stumped, (Angry) Trigonometry has proven to not be my strongest subject. Any help, or something worked out to push me in the right direction, would be fantastic!

• Apr 10th 2013, 03:48 PM
mathguy25
Re: Need help simplifying this trigonometric substitution problem on homework.
sqrt(x^2 - 49)/x where x = 7 sec(theta)
sqrt(7(sec(theta)))^2 - 49) / (7sec(theta))
sqrt(49(sec(theta))^2 - 49)/ (7sec(theta))
sqrt(49[(sec(theta))^2 - 1]) / (7sec(theta))
sqrt{49(tan(theta))^2 / (7sec(theta))
(7 tan(theta)) / (7sec(theta))
7 tan(theta) cos(theta)/ 7
tan(theta) cos(theta)
sin(theta)/cos(theta) * cos(theta)
sin(theta)
• Apr 10th 2013, 04:18 PM
Soroban
Re: Need help simplifying this trigonometric substitution problem on homework.
Hello, AKPunk49!

I'll write it out in LaTeX . . . may be easier to read.

Quote:

$\text{Simplify: }\:\frac{\sqrt{x^2-49}}{x},\:\text{ where }x = 7\sec\theta$

$\frac{\sqrt{x^2-49}}{x} \;=\;\frac{\sqrt{(7\sec\theta)^2 - 49}}{7\sec\theta} \;=\;\frac{\sqrt{49\sec^2\!\theta - 49}}{7\sec\theta}$

. . . . . . . $=\;\frac{\sqrt{49(\sec^2\!\theta-1)}}{\7\sec\theta} \;=\; \frac{\sqrt{49\tan^2\!\theta}}{7\sec\theta}$

. . . . . . . $=\;\frac{7\tan\theta}{7\sec\theta} \;=\;\frac{\tan\theta}{\sec\theta} \;=\;\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1} {\cos\theta}} \;=\;\sin\theta$