What is the value of sin^2(1/2 arccos(-4/5)) ?

The value of sin^2(1/2 arccos(-4/5)) is 9/10

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sin^2(1/2 arccos(-4/5))

sin^{2} (\frac{1}{2} arccos (- \frac{4}{5}))

\frac{9}{10}

Solution steps

sin^{2} (\frac{1}{2} arccos (- \frac{4}{5}))

Simplify:

\frac{1}{2} arccos (- \frac{4}{5}) = \frac{arccos ( - \frac{4}{5} )}{2}

= sin^{2} (\frac{arccos ( - \frac{4}{5} )}{2})

Rewrite using trig identities:

\frac{1 - cos ( arccos ( - \frac{4}{5} ) )}{2}

= \frac{1 - cos ( arccos ( - \frac{4}{5} ) )}{2}

Rewrite using trig identities:

cos (arccos (- \frac{4}{5})) = - \frac{4}{5}

= \frac{1 - ( - \frac{4}{5} )}{2}

Simplify

\frac{1 - ( - \frac{4}{5} )}{2} : \frac{9}{10}

= \frac{9}{10}

16 sin (3 8^{\circ})

cos (- 81 0^{\circ})

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

cos (38.6 6^{\circ})

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

What is the value of sin^2(1/2 arccos(-4/5)) ?
The value of sin^2(1/2 arccos(-4/5)) is 9/10