What is the value of -1/(sin^2(60)) ?

The value of -1/(sin^2(60)) is -4/3

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-1/(sin^2(60))

- \frac{1}{sin ^{2} ( 6 0 ^{\circ} )}

- \frac{4}{3}

Decimal Notation

- 1.33333 \dots

Solution steps

- \frac{1}{sin ^{2} ( 6 0 ^{\circ} )}

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= - \frac{1}{( \frac{3}{2} ) ^{2}}

Simplify

- \frac{1}{( \frac{3}{2} ) ^{2}} : - \frac{4}{3}

- \frac{1}{( \frac{3}{2} ) ^{2}}

(\frac{3}{2})^{2} = \frac{3}{2 ^{2}}

(\frac{3}{2})^{2}

Apply exponent rule:

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

= \frac{( 3 ) ^{2}}{2 ^{2}}

(3)^{2} : 3

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 3

= \frac{3}{2 ^{2}}

= - \frac{1}{\frac{3}{2 ^{2}}}

Apply the fraction rule:

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

= - \frac{2 ^{2}}{3}

2^{2} = 4

= - \frac{4}{3}

= - \frac{4}{3}

- 2 sin (2 (\frac{π}{4}))

sin (arctan (\frac{3}{2}))

2 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))

tan (33.6 9^{\circ})

\frac{tan ( 6 0 ^{\circ} ) sin ( 3 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )}