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

The value of -3cos(2(pi/3))+1 is 5/2

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-3cos(2(pi/3))+1

- 3 cos (2 (\frac{π}{3})) + 1

\frac{5}{2}

Decimal Notation

2.5

Solution steps

- 3 cos (2 (\frac{π}{3})) + 1

= - 3 cos (2 \cdot \frac{π}{3}) + 1

Rewrite using trig identities:

cos (2 \cdot \frac{π}{3}) = - \frac{1}{2}

cos (2 \cdot \frac{π}{3})

Simplify:

2 \cdot \frac{π}{3} = \frac{2 π}{3}

2 \cdot \frac{π}{3}

Multiply fractions:

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

= \frac{π 2}{3}

= cos (\frac{2 π}{3})

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 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}

= - \frac{1}{2}

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

Simplify

- 3 (- \frac{1}{2}) + 1 : \frac{5}{2}

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

Apply rule

- (- a) = a

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

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

3 \cdot \frac{1}{2}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2}

= \frac{3}{2} + 1

Convert element to fraction:

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

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

1 \cdot 2 + 3 = 5

1 \cdot 2 + 3

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 3

Add the numbers:

2 + 3 = 5

= 5

= \frac{5}{2}

= \frac{5}{2}

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

\frac{1}{cos ( \frac{1}{4} )}

9 cos (1 2^{\circ})

cos (- 20 0^{\circ})

arctan (\frac{56}{91})