Tan[arctan(2/3) + arccos(8/17)]

  • Thread starter jrjack
  • Start date
  • #1
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Homework Statement



tan[arctan(2/3)+arccos(8/17)]

Homework Equations



tan(u+v)=[itex]\frac{\tan u+\tan v}{1-\tan u\tan v}[/itex]

The Attempt at a Solution



After drawing 2 triangles for arctan and arccos (both in Quadrant 1), and inserting them into the addition formula for tan, I get:
[tex]\frac{\frac{2}{3}+\frac{15}{8}}{1-\frac{2}{3}(\frac{15}{8})}[/tex]
is that right?
if so, then maybe my math is off...
I got the numerator reduced to 23/8
and denominator of -1/4
for a total of -23/3 ...but thats not correct.
 

Answers and Replies

  • #2
Dick
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I think your math with the fractions is off. 2/3+15/8 is not equal to 23/8. Can you show why you think it is?
 
  • #3
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Sorry, mis-typed...I meant 23/12

2/3 =16/24
15/8=30/24******Found my problem, this should be 45/24


Thanks for looking at this, I knew it had to be something simple...I need to slow down.
 

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