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

The value of sin(2pi)-sin(pi) is 0

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sin(2pi)-sin(pi)

sin (2 π) - sin (π)

0

Solution steps

sin (2 π) - sin (π)

Rewrite using trig identities:

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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

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 - 0

Simplify

= 0

sin (13 5^{\circ}) + cos (13 5^{\circ}) + tan (13 5^{\circ})

tan (arcsin (\frac{- 12}{13}))

sec (- 84 0^{\circ})

sec (arccos (- \frac{3}{5}))

sin (3 3^{\circ} + 4 2^{\circ})