Trigonometry assistance

  • Thread starter TonyC
  • Start date
86
0
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?
 

TD

Homework Helper
1,020
0
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]
 
86
0
That makes perfect sense now.....

The light has come on, thank you
 

TD

Homework Helper
1,020
0
No problem :smile:
 

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