What is the value of 2cos^2(pi/3)-1 ?

The value of 2cos^2(pi/3)-1 is -1/2

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2cos^2(pi/3)-1

2 cos^{2} (\frac{π}{3}) - 1

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

2 cos^{2} (\frac{π}{3}) - 1

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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{1}{2}

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

Simplify

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

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

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

2 (\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}}

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

Cancel the common factor:

2

= \frac{1}{2}

= \frac{1}{2} - 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

= - \frac{1}{2}

- cot (\frac{3 π}{4})

cos (\frac{5 π}{4} - \frac{5 π}{6})

3 cos (3 π)

sin (2 \cdot 50)

\frac{10}{sin ( 3 0 ^{\circ} )}