What is the value of cos(-60)-tan(-45) ?

The value of cos(-60)-tan(-45) is 3/2

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cos(-60)-tan(-45)

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

Use the following property:

tan (- x) = - tan (x)

tan (- 4 5^{\circ}) = - tan (4 5^{\circ})

= cos (- 6 0^{\circ}) - (- tan (4 5^{\circ}))

Use the following property:

cos (- x) = cos (x)

cos (- 6 0^{\circ}) = cos (6 0^{\circ})

= cos (6 0^{\circ}) - (- tan (4 5^{\circ}))

Simplify

= cos (6 0^{\circ}) + tan (4 5^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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}

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\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}

= 1

= \frac{1}{2} + 1

Simplify

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

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{3}{2}

cos (arccos (\frac{2}{7}))

cos (arccot (2) + arccot (3))

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

sin (- 3 0^{\circ}) + sin (15 0^{\circ})

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