What is the value of -2cos(135) ?

The value of -2cos(135) is square root of 2

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-2cos(135)

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

2

Decimal Notation

1.41421 \dots

Solution steps

- 2 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}

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

Simplify

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

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

Apply rule

- (- a) = a

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

Multiply fractions:

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

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

Cancel the common factor:

2

= 2

= 2

3 sin (\frac{7 π}{6})

sin^{2} (22.6 2^{\circ})

13 cos (6 5^{\circ})

2 sin (2 \frac{π}{2})

tan (\frac{4}{7})