What is the value of 12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2}) ?

The value of 12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2}) is 62/5-(5(4pi-3*3^{1/2}))/6

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12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2})

12.4 - 10 (0.5 π - arcsin (0.5) - 0.5 (1 - (0.5)^{2})^{\frac{1}{2}})

\frac{62}{5} - \frac{5 ( 4 π - 3 \cdot 3 ^{\frac{1}{2}} )}{6}

Solution steps

12.4 - 10 (0.5 π - arcsin (0.5) - 0.5 (1 - (0.5)^{2})^{\frac{1}{2}})

= \frac{62}{5} - 10 \frac{1}{2} π - arcsin (\frac{1}{2}) - \frac{1}{2} (1 - (\frac{1}{2})^{2})^{\frac{1}{2}}

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

= \frac{62}{5} - 10 \frac{1}{2} π - \frac{π}{6} - \frac{1}{2} (1 - (\frac{1}{2})^{2})^{\frac{1}{2}}

Simplify

\frac{62}{5} - 10 \frac{1}{2} π - \frac{π}{6} - \frac{1}{2} (1 - (\frac{1}{2})^{2})^{\frac{1}{2}} : \frac{62}{5} - \frac{5 ( 4 π - 3 \cdot 3 ^{\frac{1}{2}} )}{6}

= \frac{62}{5} - \frac{5 ( 4 π - 3 \cdot 3 ^{\frac{1}{2}} )}{6}

sin (\frac{11}{9}) π

\frac{1.24}{tan ( 3 4 ^{\circ} )}

\frac{sin ( 3 0 ^{\circ} )}{sin ( 2 2 ^{\circ} )}

tan (21. 8^{\circ}) 2

12.1 cos (3 0^{\circ})

What is the value of 12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2}) ?
The value of 12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2}) is 62/5-(5(4pi-3*3^{1/2}))/6