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ianwalters

  • 2 years ago

verify: secx - cosx = tanxsinx

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  1. whpalmer4
    • 2 years ago
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    What's the definition of \(\sec x\)?

  2. ianwalters
    • 2 years ago
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    cosx

  3. whpalmer4
    • 2 years ago
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    No. Why would we need a different name if it is the same?

  4. ianwalters
    • 2 years ago
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    oh oops. its 1/cosx

  5. whpalmer4
    • 2 years ago
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    There we go. So what if you substitute 1/cos x in to the equation, what do you get?

  6. ianwalters
    • 2 years ago
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    1/cosx - cosx = tanxsinx

  7. whpalmer4
    • 2 years ago
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    Okay, can you make a common denominator and combine 1/cos x - cos x?

  8. ianwalters
    • 2 years ago
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    now: (1-cos^2x)/cosx

  9. whpalmer4
    • 2 years ago
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    okay, does that numerator look like any of your trig identities?

  10. ianwalters
    • 2 years ago
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    yes but i cant remember what that is equivalent to

  11. whpalmer4
    • 2 years ago
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    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 \(\sin^2x = 1-\cos^2x\)

  12. whpalmer4
    • 2 years ago
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    rats, left out a '('! \(x^2+y^2=1\)

  13. whpalmer4
    • 2 years ago
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    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!

  14. ianwalters
    • 2 years ago
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    thank you so much!

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