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whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
What's the definition of \(\sec x\)?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
No. Why would we need a different name if it is the same?
 one year ago

ianwalters Group TitleBest ResponseYou've already chosen the best response.0
oh oops. its 1/cosx
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
There we go. So what if you substitute 1/cos x in to the equation, what do you get?
 one year ago

ianwalters Group TitleBest ResponseYou've already chosen the best response.0
1/cosx  cosx = tanxsinx
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Okay, can you make a common denominator and combine 1/cos x  cos x?
 one year ago

ianwalters Group TitleBest ResponseYou've already chosen the best response.0
now: (1cos^2x)/cosx
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
okay, does that numerator look like any of your trig identities?
 one year ago

ianwalters Group TitleBest ResponseYou've already chosen the best response.0
yes but i cant remember what that is equivalent to
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
well, the formula for a circle at the origin is \(x^2 + y^2 = r^2\) or if we are talking the unit circle used to define the trig functions, \x^2 + y^2 = 1\) and so \(\sin^2x + \cos^x = 1\) or \(\sin^2x = 1\cos^2x\)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
rats, left out a '('! \(x^2+y^2=1\)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
So our left hand side is now \[\frac{\sin^2x}{\cos x}\]Wouldn't it be great if \[\tan x = \frac{\sin x}{\cos x}\] because that would make our right hand side be \[\frac{\sin x}{\cos x}*\sin x = \frac{\sin^2x}{\cos x}\]just like our left hand side. SOH CAH TOA sine = opposite over hypotenuse cosine = adjacent over hypotenuse tangent = opposite over adjacent Hey, look at that: opposite over hypotenuse divided by adjacent over hypotenuse = opposite over adjacent, so \[\tan x =\frac{\sin x}{\cos x}\]just like we want!
 one year ago

ianwalters Group TitleBest ResponseYou've already chosen the best response.0
thank you so much!
 one year ago
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