What is the value of 2sin(3pi) ?

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2sin(3pi)

2 sin (3 π)

0

Solution steps

2 sin (3 π)

Rewrite using trig identities:

sin (3 π) = 0

sin (3 π)

sin (3 π) = sin (π)

sin (3 π)

Rewrite

3 π

2 π + π

= sin (2 π + π)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + π) = sin (π)

= sin (π)

= sin (π)

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

= 0

= 2 \cdot 0

Simplify

= 0

(sin (6 0^{\circ})) (cos (3 0^{\circ})) - (sin (3 0^{\circ})) (cos (6 0^{\circ}))

sin (arctan (\frac{2}{5}))

\frac{sin ( 6 0 ^{\circ} ) \cdot tan ( 3 0 ^{\circ} )}{sec ( 4 5 ^{\circ} )}

- 4 π^{2} cos (π + \frac{π}{4})

sin (\frac{13}{12} π)