What is the value of sec(-(17pi)/4) ?

The value of sec(-(17pi)/4) is square root of 2

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sec(-(17pi)/4)

sec (- \frac{17 π}{4})

2

Decimal Notation

1.41421 \dots

Solution steps

sec (- \frac{17 π}{4})

Use the following property:

sec (- x) = sec (x)

sec (- \frac{17 π}{4}) = sec (\frac{17 π}{4})

= sec (\frac{17 π}{4})

sec (\frac{17 π}{4}) = sec (\frac{π}{4})

sec (\frac{17 π}{4})

Rewrite

\frac{17 π}{4}

2 π \cdot 2 + \frac{π}{4}

= sec (2 π 2 + \frac{π}{4})

Apply the periodicity of

sec

sec (x + 2 π \cdot k) = sec (x)

sec (2 π \cdot 2 + \frac{π}{4}) = sec (\frac{π}{4})

= sec (\frac{π}{4})

= sec (\frac{π}{4})

Use the following trivial identity:

sec (\frac{π}{4}) = 2

sec (\frac{π}{4})

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 2

= 2

sin (arcsin (\frac{7}{25}) - arccos (- \frac{5}{13}))

arccos (\frac{1}{9})

arctan (\frac{- 6}{- 3})

2 cos^{2} (\frac{5 π}{12}) - 1

8 tan (\frac{π}{4})

What is the value of sec(-(17pi)/4) ?
The value of sec(-(17pi)/4) is square root of 2