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

The value of csc((13pi)/4) is -sqrt(2)

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csc((13pi)/4)

Solution

csc (\frac{13 π}{4})

Solution

- 2

Decimal Notation

- 1.41421 \dots

Solution steps

csc (\frac{13 π}{4})

csc (\frac{13 π}{4}) = csc (\frac{5 π}{4})

csc (\frac{13 π}{4})

Rewrite

\frac{13 π}{4}

2 π + \frac{5 π}{4}

= csc (2 π + \frac{5 π}{4})

Apply the periodicity of

csc

csc (x + 2 π) = csc (x)

csc (2 π + \frac{5 π}{4}) = csc (\frac{5 π}{4})

= csc (\frac{5 π}{4})

= csc (\frac{5 π}{4})

Rewrite using trig identities:

\frac{1}{sin ( \frac{5 π}{4} )}

csc (\frac{5 π}{4})

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( \frac{5 π}{4} )}

= \frac{1}{sin ( \frac{5 π}{4} )}

Rewrite using trig identities:

sin (\frac{5 π}{4}) = - \frac{2}{2}

sin (\frac{5 π}{4})

Rewrite using trig identities:

sin (π) cos (\frac{π}{4}) + cos (π) sin (\frac{π}{4})

sin (\frac{5 π}{4})

Write

sin (\frac{5 π}{4})

sin (π + \frac{π}{4})

= sin (π + \frac{π}{4})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (π) cos (\frac{π}{4}) + cos (π) sin (\frac{π}{4})

= sin (π) cos (\frac{π}{4}) + cos (π) sin (\frac{π}{4})

Use the following trivial identity:

sin (π) = 0

sin (π)

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= 0 \cdot \frac{2}{2} + (- 1) \frac{2}{2}

Simplify

= - \frac{2}{2}

= \frac{1}{- \frac{2}{2}}

Simplify

\frac{1}{- \frac{2}{2}} : - 2

\frac{1}{- \frac{2}{2}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{\frac{2}{2}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

\frac{1}{\frac{2}{2}} = \frac{2}{2}

= - \frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= - 2

= - 2

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Frequently Asked Questions (FAQ)

What is the value of csc((13pi)/4) ?
The value of csc((13pi)/4) is -sqrt(2)