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anonymous
 one year ago
1 last trig problem
anonymous
 one year ago
1 last trig problem

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Try rewriting everything in terms of either sine or cosine

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1eg: sec(x) = 1/cos(x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1also, when it comes to proving identities, you should only work on one side

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it might be easier to work on the right side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[q \cot q/\cos^{1} \sin(\theta)/expsec/\csc ^2\frac{ f1/\sec}{ ? } \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0CRAP, i hate uploading equations on this website

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one sec, let me try again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I got my answer, but when I try to put it in the equation on this website it messes up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i started working on the right side

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1After trying out the right side, it's somewhat more complicated than I thought. So I'm going to work on the left side instead \[\Large \frac{\sec(\beta)}{\sin(\beta)}\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large {\color{red}{\frac{\sec(\beta)}{\sec(\beta)}*}}\frac{\sec(\beta)}{\sin(\beta)}\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}{\color{red}{\frac{\sin(\beta)}{\sin(\beta)}}}*\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}\frac{\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] Take a few minutes and look all that over. Then let me know when you're ready for the next step.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait I think I need to prove the right side of the equation not the left

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you can start on either side

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1just as long as you only work with one side

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1look at the "Pythagorean Identities" on page 2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what is sec^2 equal to?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1and what is sin^2 equal to?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1So, \[\Large \frac{\sec^2(\beta)\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\tan^2(\beta)+1(1\cos^2(\beta))}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1simplify the numerator for the denominator, use the fact that sin*sec = sin*(1/cos) = sin/cos = ???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I did it, can you show me what you got

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1when you simplified the numerator, what did you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wish I could show you, trust me, but I cant

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1or type what you can

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got back to where we started,

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \tan^2(\beta)+1(1\cos^2(\beta))\] \[\Large \tan^2(\beta)+11+\cos^2(\beta)\] which becomes what?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what the other side of the equation is

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah ultimately the goal is to get the left side equal to the right side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the denominator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what is sin(x)*sec(x) equal to?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1hint: look at "Tangent and Cotangent Identities" and "Reciprocal Identities" on that pdf

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1sin(x)*sec(x) = sin(x)*(1/cos(x)) = sin(x)/cos(x) = ????
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