What is the value of csc(-(11pi)/4) ?

The value of csc(-(11pi)/4) is -sqrt(2)

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csc(-(11pi)/4)

csc (- \frac{11 π}{4})

- 2

Decimal

- 1.41421 \dots

Solution steps

csc (- \frac{11 π}{4})

Use the following property:

csc (- x) = - csc (x)

csc (- \frac{11 π}{4}) = - csc (\frac{11 π}{4})

= - csc (\frac{11 π}{4})

csc (\frac{11 π}{4}) = csc (\frac{3 π}{4})

csc (\frac{11 π}{4})

Rewrite

\frac{11 π}{4}

2 π + \frac{3 π}{4}

= csc (2 π + \frac{3 π}{4})

Apply the periodicity of

csc

csc (x + 2 π) = csc (x)

csc (2 π + \frac{3 π}{4}) = csc (\frac{3 π}{4})

= csc (\frac{3 π}{4})

= - csc (\frac{3 π}{4})

Rewrite using trig identities:

csc (\frac{3 π}{4}) = 2

csc (\frac{3 π}{4})

Rewrite using trig identities:

\frac{1}{sin ( \frac{3 π}{4} )}

csc (\frac{3 π}{4})

Use the basic trigonometric identity:

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

= \frac{1}{sin ( \frac{3 π}{4} )}

= \frac{1}{sin ( \frac{3 π}{4} )}

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 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}

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

Simplify

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

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

Apply the fraction rule:

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

= \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

= - 2