anonymous
  • anonymous
can someone explain how sec(-pi/4) = sqrt2? I have never encountered a negative angle like this roflmao
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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goformit100
  • goformit100
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anonymous
  • anonymous
Are you sure it has a negative?
jdoe0001
  • jdoe0001
\(\bf sec\left(-\frac{\pi}{4}\right) = \sqrt{2}\\\quad \\ -----------------------\\ \textit{keep in mind that}\\ sec(\theta) = \cfrac{1}{cos(\theta)} \implies sec\left(-\frac{\pi}{4}\right) = \cfrac{1}{cos\left(-\frac{\pi}{4}\right)}\\ cos(-\theta) = cos(\theta) \implies \cfrac{1}{cos\left(-\frac{\pi}{4}\right)} \implies \cfrac{1}{cos\left(\frac{\pi}{4}\right)}\\\quad \\ -----------------------\\ sec\left(-\frac{\pi}{4}\right) = \sqrt{2} \implies \cfrac{1}{cos\left(\frac{\pi}{4}\right)} =\sqrt{2} \implies \cfrac{1}{\sqrt{2}} = cos\left(\frac{\pi}{4}\right)\)

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jdoe0001
  • jdoe0001
\(\bf \cfrac{1}{\sqrt{2}} \implies \cfrac{1}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}} \implies \cfrac{\sqrt{2}}{2}\\ ------------------------\\ \cfrac{\sqrt{2}}{2}= cos\left(\frac{\pi}{4}\right)\) and you can check your Unit Circle for that
jdoe0001
  • jdoe0001
|dw:1378750719345:dw|
jdoe0001
  • jdoe0001
is really the same circle, the angles just start off in different directions, for the positive ones, the ones you're used to, you start off from 0 and go UPWARDS for the negative ones, you start off on the 0 and go DOWNWARDS
jdoe0001
  • jdoe0001
come to think, you never asked for an identity conversion... hm why did I do it, dunno hehhehe, anyhow, that'd be another way to write the same identity, and that's what a negative angle is
jdoe0001
  • jdoe0001
one can also say that \(\bf sec\left(-\frac{\pi}{4}\right) = \sqrt{2}\\\quad \\ -----------------------\\ sec(\theta) = \cfrac{1}{cos(\theta)} \implies sec\left(-\frac{\pi}{4}\right) = \cfrac{1}{cos\left(-\frac{\pi}{4}\right)}\\ \implies sec\left(-\frac{\pi}{4}\right) = \cfrac{1}{cos\left(\frac{\pi}{4}\right)} = sec\left(\frac{\pi}{4}\right)\\ -----------------------\\ sec\left(-\frac{\pi}{4}\right) = \sqrt{2} \implies sec\left(\frac{\pi}{4}\right) = \sqrt{2}\)

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