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

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

whpalmer4
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.01/cosx  cosx = tanxsinx

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

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

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

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

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

whpalmer4
 2 years 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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