Inverse Trig Functions: Evaluating Expressions in Radians

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SUMMARY

The discussion focuses on evaluating inverse trigonometric functions, specifically arctan expressions in radians. The correct evaluations are: a) arctan(-(sqrt3)/3) = -π/6, b) arctan((sqrt3)/3) = π/6, and c) arctan(-sqrt3) = -π/3. Participants clarify that the principal value of arctan is typically restricted to the range of -π/2 to π/2, contrasting with the broader range of 0 to 2π used in other functions like arccos and arcsin.

PREREQUISITES
  • Understanding of inverse trigonometric functions
  • Familiarity with radians and their conversions
  • Knowledge of the tangent function and its properties
  • Basic grasp of the principal value concept in trigonometry
NEXT STEPS
  • Study the properties of inverse trigonometric functions in detail
  • Learn about the range and principal values of arctan and other inverse functions
  • Explore the relationship between tangent and its inverse functions
  • Investigate the applications of inverse trigonometric functions in real-world problems
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Mathematics students, educators, and anyone interested in mastering trigonometric functions and their applications in various fields.

shiri
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Evaluate the following expressions. Your answer must be in radians.

a) arctan(-(sqrt3)/3)

b) arctan((sqrt3)/3)

c) arctan(-sqrt3)


What I got are:

a) 5pi/6

b) pi/6

c) 2pi/3


Do I got these answers right?
 
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Hi shiri! :smile:

(have a pi: π and a square-root: √ :wink:)

Yes and no.

The tangents of those angles are the numbers given.

But "arctan" mean the principal value, which is usually taken to be between -π/2 and π/2 …

see http://en.wikipedia.org/wiki/Arctan" .

(The reason why that range is chosen (and not the [0,2π) for arccos and arcsin) is because tan goes smoothly from -∞ to ∞ in that range.)

You've chosen the range from 0 to 2π (of course, if your professor told you to do so, ignore wikipedia and me :wink:).
 
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