What is the value of 1/3 (-8cos(0)-(tan^2(0))/2) ?

The value of 1/3 (-8cos(0)-(tan^2(0))/2) is -8/3

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1/3 (-8cos(0)-(tan^2(0))/2)

\frac{1}{3} (- 8 cos (0) - \frac{tan ^{2} ( 0 )}{2})

- \frac{8}{3}

Decimal Notation

- 2.66666 \dots

Solution steps

\frac{1}{3} (- 8 cos (0) - \frac{tan ^{2} ( 0 )}{2})

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= \frac{1}{3} (- 8 \cdot 1 - \frac{0 ^{2}}{2})

Simplify

\frac{1}{3} (- 8 \cdot 1 - \frac{0 ^{2}}{2}) : - \frac{8}{3}

\frac{1}{3} (- 8 \cdot 1 - \frac{0 ^{2}}{2})

Apply rule

0^{a} = 0

0^{2} = 0

= \frac{1}{3} (- \frac{0}{2} - 8 \cdot 1)

Apply rule

\frac{0}{a} = 0, a \neq = 0

\frac{0}{2} = 0

= \frac{1}{3} (- 8 \cdot 1 - 0)

Multiply the numbers:

8 \cdot 1 = 8

= \frac{1}{3} (- 8 - 0)

Subtract the numbers:

- 8 - 0 = - 8

= (- 8) \frac{1}{3}

Remove parentheses:

(- a) = - a

= - \frac{1}{3} \cdot 8

Multiply fractions:

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

= - \frac{1 \cdot 8}{3}

Multiply the numbers:

1 \cdot 8 = 8

= - \frac{8}{3}

= - \frac{8}{3}

30 sin (5 5^{\circ})

arcsin (\frac{3 3}{6})

\frac{arcsin ( \frac{1}{3} ) - arcsin ( - \frac{1}{3} )}{600}

450 sin (7 5^{\circ})

sinh (\frac{7}{17})

What is the value of 1/3 (-8cos(0)-(tan^2(0))/2) ?
The value of 1/3 (-8cos(0)-(tan^2(0))/2) is -8/3