What is the value of -2sin(1-1) ?

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-2sin(1-1)

- 2 sin (1 - 1)

0

Solution steps

- 2 sin (1 - 1)

Rewrite using trig identities:

sin (1 - 1) = 0

sin (1 - 1)

Simplify

= 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

= - 2 \cdot 0

Simplify

= 0

50 \cdot sin (1 5^{\circ})

\frac{sin ( 15 0 ^{\circ} ) + sin ( 12 0 ^{\circ} )}{cos ( 20 0 ^{\circ} ) - cos ( 30 0 ^{\circ} )}

4.03 sin (2 π) (\frac{200 - 76}{365}) + 11.43

arctan (\frac{142.14}{140.31})

0.302 \cdot sin (6 0^{\circ})