anonymous
  • anonymous
1 last trig problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
Try rewriting everything in terms of either sine or cosine
jim_thompson5910
  • jim_thompson5910
eg: sec(x) = 1/cos(x)

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anonymous
  • anonymous
ahhhhh ok
anonymous
  • anonymous
give me one sec
jim_thompson5910
  • jim_thompson5910
also, when it comes to proving identities, you should only work on one side
jim_thompson5910
  • jim_thompson5910
it might be easier to work on the right side
anonymous
  • anonymous
\[q \cot q/\cos^{-1} -\sin(\theta)/expsec/\csc ^2\frac{ f1/\sec}{ ? } \]
anonymous
  • anonymous
CRAP, i hate uploading equations on this website
anonymous
  • anonymous
one sec, let me try again
anonymous
  • anonymous
ok I got my answer, but when I try to put it in the equation on this website it messes up
anonymous
  • anonymous
so i started working on the right side
jim_thompson5910
  • jim_thompson5910
After 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
  • anonymous
wait I think I need to prove the right side of the equation not the left
jim_thompson5910
  • jim_thompson5910
you can start on either side
anonymous
  • anonymous
ok i get it
jim_thompson5910
  • jim_thompson5910
just as long as you only work with one side
anonymous
  • anonymous
alright is that it
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
look at the "Pythagorean Identities" on page 2
jim_thompson5910
  • jim_thompson5910
what is sec^2 equal to?
anonymous
  • anonymous
tan^2(theta)+1
jim_thompson5910
  • jim_thompson5910
and what is sin^2 equal to?
anonymous
  • anonymous
1-cos^2(theta)
jim_thompson5910
  • jim_thompson5910
So, \[\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\tan^2(\beta)+1-(1-\cos^2(\beta))}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
jim_thompson5910
  • jim_thompson5910
simplify the numerator for the denominator, use the fact that sin*sec = sin*(1/cos) = sin/cos = ???
anonymous
  • anonymous
Ok I did it, can you show me what you got
jim_thompson5910
  • jim_thompson5910
when you simplified the numerator, what did you get?
anonymous
  • anonymous
i wish I could show you, trust me, but I cant
jim_thompson5910
  • jim_thompson5910
draw it out
jim_thompson5910
  • jim_thompson5910
or type what you can
anonymous
  • anonymous
I got back to where we started,
jim_thompson5910
  • jim_thompson5910
\[\Large \tan^2(\beta)+1-(1-\cos^2(\beta))\] \[\Large \tan^2(\beta)+1-1+\cos^2(\beta)\] which becomes what?
anonymous
  • anonymous
what the other side of the equation is
jim_thompson5910
  • jim_thompson5910
yeah ultimately the goal is to get the left side equal to the right side
anonymous
  • anonymous
what about the denominator
jim_thompson5910
  • jim_thompson5910
what is sin(x)*sec(x) equal to?
jim_thompson5910
  • jim_thompson5910
hint: look at "Tangent and Cotangent Identities" and "Reciprocal Identities" on that pdf
anonymous
  • anonymous
ok I got it
anonymous
  • anonymous
it is sinxsec
jim_thompson5910
  • jim_thompson5910
sin(x)*sec(x) = sin(x)*(1/cos(x)) = sin(x)/cos(x) = ????
anonymous
  • anonymous
tan
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
alright thanks

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