0 therefore to solve tan θ a you must know which of

This preview shows page 8 - 11 out of 14 pages.

0. Therefore, to solve tanθ=a, you must knowwhich of the two possible quadrants the angle is in and then make a correction if the angle isin the second or third quadrants. Fortunately, the correction is easy to remember: “just addπTo solve tanθ=a, these are the four possibilities.(1) tanθ=a,a >0 andθis in the first quadrant=θ= arctana(2) tanθ=a,a <0 andθis in the second quadrant=θ= arctana+π(3) tanθ=a,a >0 andθis in the third quadrant=θ= arctana+π(4) tanθ=a,a <0 andθis in the fourth quadrant=(θ= arctanaorθ= (arctana) + 2π
Once again, in tabular form, to solve:tan θ=aIfaisandθis inthenθequalspositivefirst quadrantarctananegativesecond quadrant(arctana) +πpositivethird quadrant(arctana) +πnegativefourth quadrantarctanaor(arctana) + 2πThus, it appears that one can not uniquely solve tanθ=aunless it is known which quadrantθis in.This is correct.Fortunately, in applications when the equation needs to be solved,information will always be available to determine which possible angle is correct.How should one remember these results? Rather simple: ifθis in the second or third quadrants,just addπradians to the computer’s evaluation of arctana.While this alone is sufficient to get a correct answer, there are two simplifications sometimesused.First of all, the function arctan is anoddfunction, i.e., arctan(-b) =-arctan(b) for anyreal numberb. This comes into play especially in the second and fourth quadrants, where thetangent is negative. So, for example, if tanθ=-1.5, thenθcan be written asθ=-arctan(1.5)if it is in the fourth quadrant, orθ=π-arctan(1.5) if is is in the second quadrant. Noticethat we have followed exactly the rules above, except we have pulled the minus sign out of thearctangent function.The second simplification refers only to fourth quadrant angles. Although the computer willgive fourth quadrant angles as negative angles when evaluating the arctangent function, insome applications it will be desirable to haveθalways between 0 and 2πradians. In this case,only for fourth quadrant angles, one will need to add 2πto obtain the same angle writtenas a positive angle.If you have understood all of this, you will realize that the solution oftanθ=-1.2, whenθis known to be in the fourth quadrant, can be written three ways: (i)θ= arctan(-1.2); (ii)θ=-arctan(1.2) , which give exactly the sameθas the first way; (iii)θ= 2π+ arctan(-1.2) = 2π-arctan(1.2) , which givesθas a positive angle.IdentitiesThe three most important trig identities (and, we believe, the ones every scientist and engineershould know) are the following.(1)sin2x+ cos2x= 1 (True for anyx; for example, sin2et+ cos2et= 1)
(2)sin(A±B) = sinAcosB±cosAsinB(3)cos(A±B) = cosAcosBsinAsinBMany of the other trig identities can be derived easily from these. For example, dividing (1)by cos2xgivestan2x+ 1 = sec2xAngleIn general, the angleθbetween the positivex-axis and the line segment from the origin to thepoint (a, b) – always measured in the counterclockwise direction – satisfiestanθ=ba.However, one can not simply concludeθ= arctanbawithout also checking quadrants.Example: Find the angle between thex-axis and line segment from (0,0) to (5,2).The answer isθ= tan-1(2

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 14 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Spring
Professor
DONTREMEMBER
Tags
Algebra, Geometry, Trigonometry, Equations, Parametric Equations, Quadratic equation, Elementary algebra

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture