Difference between revisions of "Relationship between arctan and arccot"

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(Proof)
 
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==Proof==
 
==Proof==
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Let $y = \arctan \left( \dfrac{1}{z} \right)$. Then since arctan is the [[inverse function]] of [[tangent]],
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$$\tan(y)=\dfrac{1}{z}.$$
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By the definition of cotangent, we get
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$$\cot(y)=z.$$
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Since $\mathrm{arccot}$ is the inverse function of $\cot$, take the $\mathrm{arccot}$ of each side to get
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$$y = \mathrm{arccot}(z).$$
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Therefore we have shown
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$$\arctan \left( \dfrac{1}{z} \right) = \mathrm{arccot}(z),$$
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as was to be shown.
  
 
==References==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 14:45, 25 September 2016

Theorem

The following formula holds: $$\mathrm{arctan}(z) = \mathrm{arccot}\left( \dfrac{1}{z} \right),$$ where $\mathrm{arctan}$ denotes the inverse tangent and $\mathrm{arccot}$ denotes the inverse cotangent.

Proof

Let $y = \arctan \left( \dfrac{1}{z} \right)$. Then since arctan is the inverse function of tangent, $$\tan(y)=\dfrac{1}{z}.$$ By the definition of cotangent, we get $$\cot(y)=z.$$ Since $\mathrm{arccot}$ is the inverse function of $\cot$, take the $\mathrm{arccot}$ of each side to get $$y = \mathrm{arccot}(z).$$ Therefore we have shown $$\arctan \left( \dfrac{1}{z} \right) = \mathrm{arccot}(z),$$ as was to be shown.

References