dolphinandwhale dolphinandwhale
  • 03-02-2020
  • Mathematics
contestada

What is the exact value of arccos(−2√2) ?



Enter your answer, in terms of π , in the box.

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kudzordzifrancis
kudzordzifrancis kudzordzifrancis
  • 10-02-2020

Answer:

See explanation

Step-by-step explanation:

The range of the cosine function is [tex]-1\le \cos x\le 1[/tex]

Therefore [tex]arc \cos (-2\sqrt{2})[/tex] is not defined.

Assuming your question is rather [tex]arc \cos (-\frac{\sqrt{2}}{2})[/tex]

Then [tex]arc \cos (-\frac{\sqrt{2}}{2})=\pi-arc \cos (\frac{\sqrt{2}}{2})[/tex]  in the second quadrant.

[tex]arc \cos (-\frac{\sqrt{2}}{2})=\pi-\frac{\pi}{4}[/tex]

[tex]arc \cos (-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}[/tex]

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