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

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

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

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

ianwalters
 one year ago
Best ResponseYou've already chosen the best response.01/cosx  cosx = tanxsinx

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

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

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

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

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2well, 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 \(\sin^2x = 1\cos^2x\)

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

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2So 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!
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