Trigonometry assistance

  • Thread starter TonyC
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  • #1
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Main Question or Discussion Point

How do I go about finding the equivalent expression for cot 50 degrees?

When I work the problem, I come up with tan 130 degrees.

Where am I going wrong?
 

Answers and Replies

  • #2
TD
Homework Helper
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I rather work in radians, if you don't mind.

[tex]50^\circ = \frac{{5\pi }}{{18}}[/tex]

We know that:
[tex]\cot \left( \alpha \right) = \tan \left( {\frac{\pi }{2} - \alpha } \right)[/tex]

So:
[tex]\cot \left( {\frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{\pi }{2} - \frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{{2\pi }}{9}} \right) = \tan \left( {40^\circ } \right)[/tex]
 
  • #3
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That makes perfect sense now.....

The light has come on, thank you
 
  • #4
TD
Homework Helper
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No problem :smile:
 

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