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Secant and Cosecant Functions

Main definitions:

  • The secant function is defined by
    secθ = 1
    cosθ
    for values of θ where cosθ ≠ 0. (For values of θ where cosθ = 0, the secant is undefined.)
  • The cosecant function is defined by
    cscθ = 1
    sinθ
    for values of θ where sinθ ≠ 0. (For values of θ where sinθ = 0, the cosecant is undefined.)

Example 1:

Draw a graph of the secant function. Label all zeroes, maxes, mins, and asymptotes. What is the period of the secant function?

Example 2:

Find the values of secant and cosecant at the common angles in the first quadrant: 0, (π/6), (π /4),)(π/3), (π/2). Determine in which other quadrants the secant and cosecant are positive.

Example 3:

Determine if the secant and cosecant functions are odd, even, or neither.

Example 4:

Draw a graph of the cosecant function. Label all zeroes, maxes, mins, and asymptotes. What is the period of the cosecant function?

Example 5:

Find the secants and cosecants of (5π/6)R, 240° , and (7π/4)R.

Find arcsin [ sin([(3π)/4]) ]

  • Plot the point [(3π)/4] on the unit circle. Note: [(3π)/4] = 135°
  • Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
  • Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arcsin[ sin([(3π)/4]) ] = [(π)/4]

Find arctan[ tan([(2π)/3]) ]

  • Plot the point [(2π)/3] on the unit circle. Note: [(2π)/3] = 120°
  • Recall that tangent is [x/y] . Now find the angle that has the same tangent value of [(2π)/3] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arctan ≤ [(π)/2]
  • Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arctan [ tan([(2π)/3]) ] = − [(π)/3]

Find arccos[ cos( − [(2π)/3]) ]

  • Plot the point − [(2π)/3] on the unit circle. Note: − [(2π)/3] = − 120°
  • Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(2π)/3] between 0 and π because 0 ≤ arccos
  • Draw a line across the x - axis to locate a corresponding point that is between 0 and π

arccos[ cos( − [(2π)/3]) ] = [(2π)/3]

Find arcsin[ sin([(5π)/6]) ]

  • Plot the point [(5π)/6] on the unit circle. Note: [(5π)/6] = 150°
  • Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(5π)/6] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arcsin ≤ [(π)/2]
  • Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arcsin[ sin([(5π)/6]) ] = [(π)/6]

Find arccos[ cos( − [(π)/6]) ]

  • Plot the point − [p/6] on the unit circle. Note: − [(π)/6] = − 30°
  • Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/6] between 0 and π because 0 ≤ arccos ≤ π
  • Draw a line across the x - axis to locate a corresponding point that is between 0 and π

arccos[ cos( − [(π)/6]) ] = [(π)/6]

Find arctan[ tan([(4π)/3]) ]

  • Plot the point [(4π)/3] on the unit circle. Note: [(4π)/3] = 240°
  • Recall that tangent is [x/y]. Now find the angle that has the same tangent value of [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
  • Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arctan[ tan([(4π)/3]) ] = [(π)/3]

Find arcsin[ sin([(5π)/4]) ]

  • Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225°
  • Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(5π)/4] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arcsin ≤ [(π)/2]
  • Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arcsin[ sin([(5π)/4]) ] = − [(π)/4]

Find arccos[ cos([(5π)/3]) ]

  • Plot the point [(5π)/3] on the unit circle. Note: [(5π)/3] = 300°
  • Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [(5π)/3] between 0 and π because 0 ≤ arccos ≤ π
  • Draw a line across the x - axis to locate a corresponding point that is between 0 and π

arccos[ cos([(5π)/3]) ] = [(π)/3]

Find arctan[ tan( − [(7π)/6]) ]

  • Plot the point − [(7π)/6] on the unit circle. Note: − [(7π)/6] = 210°
  • Recall that tangent is [x/y]. Now find the angle that has the same tangent value of − [(7π)/6] between − [(π)/2] and [(π)/2] because - [(π)/2] ≤ arctan ≤ [(π)/2]
  • Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arctan[ tan( − [(7π)/6]) ] = − [(π)/6]

Find arcsin[ sin( − [(5π)/6]) ]

  • Plot the point − [(5π)/6] on the unit circle. Note: − [(5π)/6] = 150°
  • Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(5π)/6] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
  • Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]

arcsin[ sin( − [(5π)/6]) ] = − [(π)/6]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Secant and Cosecant Functions

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Mathematics: Trigonometry

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