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

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SUMMARY

The discussion centers on the calculation of tan[arctan(2/3) + arccos(8/17)] using the tangent addition formula. The formula applied is tan(u+v) = (tan u + tan v) / (1 - tan u * tan v). The user initially miscalculated the numerator and denominator but later corrected the numerator to 45/24, indicating a need for careful fraction addition. The final result of the tangent calculation is confirmed to be -23/3, highlighting the importance of accuracy in mathematical operations.

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  • Understanding of trigonometric functions, specifically arctan and arccos.
  • Familiarity with the tangent addition formula.
  • Basic skills in fraction addition and simplification.
  • Knowledge of triangle properties in trigonometry.
NEXT STEPS
  • Study the derivation and applications of the tangent addition formula.
  • Practice solving problems involving arctan and arccos functions.
  • Learn techniques for accurate fraction addition and simplification.
  • Explore the properties of triangles in different quadrants for trigonometric functions.
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Homework Statement



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

Homework Equations



tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}

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:
\frac{\frac{2}{3}+\frac{15}{8}}{1-\frac{2}{3}(\frac{15}{8})}
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 that's not correct.
 
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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?
 


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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