What is the value of cos(5pi)+isin(5pi) ?

The value of cos(5pi)+isin(5pi) is -1

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cos(5pi)+isin(5pi)

cos (5 π) + i sin (5 π)

- 1

Solution steps

cos (5 π) + i sin (5 π)

cos (5 π) = cos (π)

cos (5 π)

Rewrite

5 π

2 π \cdot 2 + π

= cos (2 π 2 + π)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 2 + π) = cos (π)

= cos (π)

= cos (π) + i sin (5 π)

sin (5 π) = sin (π)

sin (5 π)

Rewrite

5 π

2 π \cdot 2 + π

= sin (2 π 2 + π)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 2 + π) = sin (π)

= sin (π)

= cos (π) + i sin (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)

Use the following trivial identity:

sin (π) = 0

sin (π)

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

= 0

= - 1 + i 0

Simplify

- 1 + i 0 : - 1

- 1 + i 0

Apply rule

0 \cdot a = 0

= - 1 + 0

Simplify

= - 1

= - 1

\frac{sin ( 15 0 ^{\circ} ) + cot ( 31 5 ^{\circ} )}{3 cos ( 21 0 ^{\circ} )}

0.23 \cdot sin (62. 2^{\circ})

tan (\frac{3 π}{16})

19.6 cos (3 0^{\circ})

tan (arccot (\frac{2 π}{3}))