What is the value of sin(2pi+pi/2) ?

The value of sin(2pi+pi/2) is 1

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sin(2pi+pi/2)

sin (2 π + \frac{π}{2})

1

Solution steps

sin (2 π + \frac{π}{2})

Simplify:

2 π + \frac{π}{2} = \frac{5 π}{2}

2 π + \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} + \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 + π}{2}

2 π 2 + π = 5 π

2 π 2 + π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π + π

Add similar elements:

4 π + π = 5 π

= 5 π

= \frac{5 π}{2}

= sin (\frac{5 π}{2})

sin (\frac{5 π}{2}) = sin (\frac{π}{2})

sin (\frac{5 π}{2})

Rewrite

\frac{5 π}{2}

2 π + \frac{π}{2}

= sin (2 π + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + \frac{π}{2}) = sin (\frac{π}{2})

= sin (\frac{π}{2})

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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}

= 1

= 1

67 sin (4 5^{\circ})

- 2 sin (3 0^{\circ})

cos (- 3)

sin (9 0^{\circ}) + cos (6 0^{\circ})

35 tan (0) - 15.022 sec^{2} (0) + 1