matthewmurphyable
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100 points!To solve the problem cos-1(cos(-pi/6)), find the angle in the interval (0,pi) whose cosine is sqrt3/2.

Sagot :

[tex]cos^{-1}[cos(\omega)]\implies \omega \\\\[-0.35em] ~\dotfill\\\\ cos\left( -\frac{\pi }{6} \right)\implies \stackrel{symmetry~identity}{cos\left( \frac{\pi }{6} \right)} \\\\\\ cos^{-1}\left[ cos\left( -\frac{\pi }{6} \right) \right]\implies cos^{-1}\left[ cos\left( \frac{\pi }{6} \right) \right]\implies \cfrac{\pi }{6}[/tex]

why did we use the positive version of π/6?

well, the inverse cosine function has a range of [0 , π], and -π/6 is on the IV Quadrant, out of the range for it, however it has a twin due to symmetry on the I Quadrant, that is π/6, thus the reason.

tingsung

Answer:

Pi over 6.

Step-by-step explanation:

I did it, but can't figure out how to add a picture.

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