What is the value of cos(120)+tan^2(240)-5 ?

The value of cos(120)+tan^2(240)-5 is -5/2

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cos(120)+tan^2(240)-5

cos (12 0^{\circ}) + tan^{2} (24 0^{\circ}) - 5

- \frac{5}{2}

Decimal Notation

- 2.5

Solution steps

cos (12 0^{\circ}) + tan^{2} (24 0^{\circ}) - 5

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} 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}

Rewrite using trig identities:

tan (24 0^{\circ}) = 3

tan (24 0^{\circ})

tan (24 0^{\circ}) = tan (6 0^{\circ})

tan (24 0^{\circ})

Rewrite

24 0^{\circ}

18 0^{\circ} + 6 0^{\circ}

= tan (18 0^{\circ} + 6 0^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 6 0^{\circ}) = tan (6 0^{\circ})

= tan (6 0^{\circ})

= tan (6 0^{\circ})

Use the following trivial identity:

tan (6 0^{\circ}) = 3

tan (6 0^{\circ})

tan (x)

periodicity table with

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

= 3

= 3

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

Simplify

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

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

(3)^{2} = 3

(3)^{2}

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{1}{2} + 3 - 5

Add/Subtract the numbers:

3 - 5 = - 2

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

Convert element to fraction:

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

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

- 2 \cdot 2 - 1 = - 5

- 2 \cdot 2 - 1

Multiply the numbers:

2 \cdot 2 = 4

= - 4 - 1

Subtract the numbers:

- 4 - 1 = - 5

= - 5

= \frac{- 5}{2}

Apply the fraction rule:

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

= - \frac{5}{2}

= - \frac{5}{2}

\frac{cos ( 6 0 ^{\circ} )}{cos ( 4 5 ^{\circ} )}

arctan (tan (\frac{5 π}{7}))

sin^{3} (1)

sin (\frac{π ^{2}}{9})

\frac{12}{sin ( 4 5 ^{\circ} )}

What is the value of cos(120)+tan^2(240)-5 ?
The value of cos(120)+tan^2(240)-5 is -5/2