What is the value of cot(-(13pi)/4) ?

The value of cot(-(13pi)/4) is -1

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cot(-(13pi)/4)

cot (- \frac{13 π}{4})

- 1

Solution steps

cot (- \frac{13 π}{4})

Use the following property:

cot (- x) = - cot (x)

cot (- \frac{13 π}{4}) = - cot (\frac{13 π}{4})

= - cot (\frac{13 π}{4})

cot (\frac{13 π}{4}) = cot (\frac{π}{4})

cot (\frac{13 π}{4})

Rewrite

\frac{13 π}{4}

π \cdot 3 + \frac{π}{4}

= cot (π 3 + \frac{π}{4})

Apply the periodicity of

cot

cot (x + π \cdot k) = cot (x)

cot (π \cdot 3 + \frac{π}{4}) = cot (\frac{π}{4})

= cot (\frac{π}{4})

= - cot (\frac{π}{4})

Use the following trivial identity:

cot (\frac{π}{4}) = 1

cot (\frac{π}{4})

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= 1

= - 1

1 - 2 sin^{2} (\frac{3 π}{7})

sec^{2} (7 0^{\circ})

arcsin (\frac{π}{8})

\frac{sec ^{2} ( \frac{π}{4} )}{2 \frac{π}{4}}

sin (\frac{9}{12})