What is the value of 4sin^2(pi/6)-1 ?

The value of 4sin^2(pi/6)-1 is 0

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4sin^2(pi/6)-1

4 sin^{2} (\frac{π}{6}) - 1

0

Solution steps

4 sin^{2} (\frac{π}{6}) - 1

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}

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

Simplify

4 (\frac{1}{2})^{2} - 1 : 0

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

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

4 (\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}}

= 4 \cdot \frac{1}{2 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 4}{2 ^{2}}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2 ^{2}}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel the common factor:

2^{2}

= 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

= 0

2 (cos (\frac{π}{6}) + i sin (\frac{π}{6})) \cdot 5 (cos (\frac{π}{3}) + i sin (\frac{π}{3}))

185 sin (1 5^{\circ})

\frac{1}{sin ( 12 0 ^{\circ} )}

cos (\frac{4 π}{3}) + i \cdot sin (\frac{4 π}{3})

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