Tan^-1 to PI Form: Get Help Here

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SUMMARY

The discussion centers on the calculation of the arctangent of the expression tan^-1(1/-√3). The user initially arrives at -30 degrees, but the correct answer, as per the revision book, is 5π/6 radians. The principal value of arctan is confirmed to lie between -π/2 and π/2. It is clarified that both -π/6 and 5π/6 yield the same tangent value, reinforcing the periodic nature of the tangent function.

PREREQUISITES
  • Understanding of trigonometric functions, specifically tangent and arctangent.
  • Familiarity with radians and degrees conversion.
  • Knowledge of De Moivre's theorem and its applications.
  • Ability to use a scientific calculator in radian mode.
NEXT STEPS
  • Study the properties of the tangent function and its periodicity.
  • Learn about the principal values of inverse trigonometric functions.
  • Explore De Moivre's theorem in depth, including its proofs and applications.
  • Practice converting between degrees and radians in trigonometric calculations.
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone seeking to deepen their understanding of inverse trigonometric functions and their applications in complex numbers.

axer
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Hello, so while i was doing my own revision. When putting tan^-1(1/-√3) i receive the answer as -30. however, (im doing demovoier's theorem) in the revision book it says the answer should be 5π/6 .

Help to brighten me? thanks.
 
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I think you are right because the principal value of arctan lies from -π/2 to π/2...
 
axer said:
Hello, so while i was doing my own revision. When putting tan^-1(1/-√3) i receive the answer as -30. however, (im doing demovoier's theorem) in the revision book it says the answer should be 5π/6 .

Help to brighten me? thanks.
make the calculation again but this time set the calculator in radians =D
if you give the answer -pi/6 don't worry is the same as 5pi/6, because tan(-pi/6)=tan(5pi/6)=1/-√3=-√3/3
 

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