anonymous
  • anonymous
verify: secx - cosx = tanxsinx
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
What's the definition of \(\sec x\)?
anonymous
  • anonymous
cosx
whpalmer4
  • whpalmer4
No. Why would we need a different name if it is the same?

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anonymous
  • anonymous
oh oops. its 1/cosx
whpalmer4
  • whpalmer4
There we go. So what if you substitute 1/cos x in to the equation, what do you get?
anonymous
  • anonymous
1/cosx - cosx = tanxsinx
whpalmer4
  • whpalmer4
Okay, can you make a common denominator and combine 1/cos x - cos x?
anonymous
  • anonymous
now: (1-cos^2x)/cosx
whpalmer4
  • whpalmer4
okay, does that numerator look like any of your trig identities?
anonymous
  • anonymous
yes but i cant remember what that is equivalent to
whpalmer4
  • whpalmer4
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 \(\sin^2x = 1-\cos^2x\)
whpalmer4
  • whpalmer4
rats, left out a '('! \(x^2+y^2=1\)
whpalmer4
  • whpalmer4
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!
anonymous
  • anonymous
thank you so much!

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