What is the value of -1/(sin^2(pi/6)) ?

The value of -1/(sin^2(pi/6)) is -4

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-1/(sin^2(pi/6))

- \frac{1}{sin ^{2} ( \frac{π}{6} )}

- 4

Solution steps

- \frac{1}{sin ^{2} ( \frac{π}{6} )}

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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{1}{2}

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

Simplify

- \frac{1}{( \frac{1}{2} ) ^{2}} : - 4

- \frac{1}{( \frac{1}{2} ) ^{2}}

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

(\frac{1}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= - 2^{2}

2^{2} = 4

= - 4

= - 4

2.4 cos (3 0^{\circ})

arcsin (sin (\frac{π}{16}))

- sec^{2} (\frac{π}{4}) \cdot tan^{2} (\frac{π}{4})

sin (4 \frac{2 π}{3})

arctan (\frac{( 15.7 - 31.83 )}{106})