Question

can someone explain how sec(-pi/4) = sqrt2? I have never encountered a negative angle like this roflmao

Answer 1

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Answer 2

Are you sure it has a negative?

Answer 3

\(\bf sec\left(-\frac{\pi}{4}\right) = \sqrt{2}\\\quad \\
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\textit{keep in mind that}\\
sec(\theta) = \cfrac{1}{cos(\theta)} \implies sec\left(-\frac{\pi}{4}\right) = \cfrac{1}{cos\left(-\frac{\pi}{4}\right)}\\
cos(-\theta) = cos(\theta) \implies \cfrac{1}{cos\left(-\frac{\pi}{4}\right)} \implies \cfrac{1}{cos\left(\frac{\pi}{4}\right)}\\\quad \\
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sec\left(-\frac{\pi}{4}\right) = \sqrt{2} \implies \cfrac{1}{cos\left(\frac{\pi}{4}\right)} =\sqrt{2} \implies \cfrac{1}{\sqrt{2}} = cos\left(\frac{\pi}{4}\right)\)

Answer 4

\(\bf \cfrac{1}{\sqrt{2}} \implies \cfrac{1}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}} \implies \cfrac{\sqrt{2}}{2}\\
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\cfrac{\sqrt{2}}{2}= cos\left(\frac{\pi}{4}\right)\)

and you can check your Unit Circle for that