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# cot(sin^-1(2/3))

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cot(sin^-1(2/3))

Guest Jun 13, 2014

#2
+92198
+5

I always draw a right angled triangle when I answer questions like this.

CPhill's answer is approximate.  It is easy to get the exact answer

Let the angle be θ

for starters sinθ is positive (1st or 2nd quad) so cotθ can be positive or neg.

From the diagram it can be seen that Cot θ= sqrt(5)/2

$$cot(sin^{-1}\left(\frac{2}{3}\right))=\frac{\sqrt{5}}{2}$$

Melody  Jun 14, 2014
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#1
+85646
+5

cot(sin^-1(2/3))

First, let's evaluate sin^-1 (2/3) = 41.810314895779°

So..........cot(41.810314895779) = 1.118033988749441

Notice that if this angle could also lie in the second quadrant..... there, the cotangent would be........ (-1.118033988749441)

CPhill  Jun 13, 2014
#2
+92198
+5

I always draw a right angled triangle when I answer questions like this.

CPhill's answer is approximate.  It is easy to get the exact answer

Let the angle be θ

for starters sinθ is positive (1st or 2nd quad) so cotθ can be positive or neg.

From the diagram it can be seen that Cot θ= sqrt(5)/2

$$cot(sin^{-1}\left(\frac{2}{3}\right))=\frac{\sqrt{5}}{2}$$

Melody  Jun 14, 2014

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