What is the value of sin(arccos(0)-arccos(1/2)) ?

The value of sin(arccos(0)-arccos(1/2)) is 1/2

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sin(arccos(0)-arccos(1/2))

sin (arccos (0) - arccos (\frac{1}{2}))

\frac{1}{2}

Solution steps

sin (arccos (0) - arccos (\frac{1}{2}))

Rewrite using trig identities:

arccos (0) - arccos (\frac{1}{2}) = arccos (\frac{3}{2})

= sin (arccos (\frac{3}{2}))

Rewrite using trig identities:

sin (arccos (\frac{3}{2})) = 1 - (\frac{3}{2})^{2}

= 1 - (\frac{3}{2})^{2}

Simplify

= \frac{1}{2}

10 tan (2 5^{\circ})

tan (3 0^{\circ} + 4 5^{\circ})

2 π cos (2 π)

arcsin (sin (3))

12 \cdot sin (4 0^{\circ})

What is the value of sin(arccos(0)-arccos(1/2)) ?
The value of sin(arccos(0)-arccos(1/2)) is 1/2