What is the value of sqrt(4cos(pi)) ?

The value of sqrt(4cos(pi)) is 2i

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sqrt(4cos(pi))

4 cos (π)

2 i

Solution steps

4 cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

= 4 (- 1)

Simplify

4 (- 1) : 2 i

4 (- 1)

Remove parentheses:

(- a) = - a

= - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= - 4

Apply radical rule:

- a = - 1 a

- 4 = - 1 4

= - 1 4

Apply imaginary number rule:

- 1 = i

= 4 i

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 i

= 2 i

arctan (0.853)

cos (4 5^{\circ}) \cdot 20

arcsin (0.314)

arcsin (0.317)

cos (\frac{13 π}{36}) cos (\frac{π}{9}) + sin (\frac{13 π}{36}) sin (\frac{π}{9})