anonymous
  • anonymous
evaluate without using a calculator cot^-1(1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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abb0t
  • abb0t
remember, cot(x) = \(\frac{cos}{sin}\)
anonymous
  • anonymous
does it matter which part i put for the inverse? Like can I put inverse cos/sin or cos/inverse sin?
anonymous
  • anonymous
inverse of cot is just 1/cot

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anonymous
  • anonymous
or sin/ cos
anonymous
  • anonymous
but 1/cot is easier for you here
anonymous
  • anonymous
is this in radians?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
How would I find the answer without using a calculator? Inverse of cot(1)
anonymous
  • anonymous
\[ \cot^{-1}(1) =x \]We can takes the \(\cot\) of both sides:\[ \cot(x) = 1 \]Then we can remember \( \cot(x) = \cos(x)/\sin(x) \)\[ \frac{\cos(x)}{\sin(x)} = 1 \]Finally we get\[ \cos(x) = \sin(x) \]
anonymous
  • anonymous
i think inverse of cotangent(1) is also equal to inverse of cotangent (adjacent side/opposite side) of a right triangle which means the adjacent side and opposite side have the same lengths. In order to have the same lengths for a right triangle the angle should be 45 degrees since it is a right triangle
anonymous
  • anonymous
Looking on the unit circle helps up imagine what this equation means: |dw:1378002537095:dw|
anonymous
  • anonymous
so \[\cot ^{-1}1=45^{o} or \pi/4 rads\]
anonymous
  • anonymous
These angles are exactly halfway between the quadrants. It is not that hard to realize the solutions. |dw:1378002672982:dw|
anonymous
  • anonymous
However, even these solutions are incomplete. You can rotate a full circle. You can continue to rotate any number of times and still be at a correct angle. |dw:1378002876690:dw|
anonymous
  • anonymous
THANK YOU SO MUCH
anonymous
  • anonymous
I UNDERSTAND NOW
anonymous
  • anonymous
Thus the complete solution set is the particular solution, and the extended solution. \[ \frac \pi 4+n\pi \quad n\in \mathbb{Z} \]
anonymous
  • anonymous
Thank you too Yeyenunez!
anonymous
  • anonymous
:)

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