What is the value of 3cos(135) ?

The value of 3cos(135) is -(3sqrt(2))/2

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3cos(135)

3 cos (13 5^{\circ})

- \frac{3 2}{2}

Decimal Notation

- 2.12132 \dots

Solution steps

3 cos (13 5^{\circ})

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 5^{\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{2}{2}

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

Simplify

3 (- \frac{2}{2}) : - \frac{3 2}{2}

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

Remove parentheses:

(- a) = - a

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

Multiply fractions:

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

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

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{3}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

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

Apply radical rule:

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

= - \frac{3}{2}

Rationalize

- \frac{3}{2} : - \frac{3 2}{2}

- \frac{3}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{3 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{3 2}{2}

= - \frac{3 2}{2}

= - \frac{3 2}{2}

1 - 2 sin^{2} (\frac{π}{7})

sin (\frac{π}{3}) cos (\frac{π}{4}) - cos (\frac{π}{3}) sin (\frac{π}{4})

tan (5 3^{\circ}) \cdot 24

arcsin (0.27)

arcsin (0.22)