What is the value of arcsin(1/2)+(1/2)/(sqrt(1-(1/2)^2)) ?

The value of arcsin(1/2)+(1/2)/(sqrt(1-(1/2)^2)) is pi/6+(sqrt(3))/3

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arcsin(1/2)+(1/2)/(sqrt(1-(1/2)^2))

arcsin (\frac{1}{2}) + \frac{\frac{1}{2}}{1 - ( \frac{1}{2} ) ^{2}}

\frac{π}{6} + \frac{3}{3}

Decimal Notation

63.08

Solution steps

arcsin (\frac{1}{2}) + \frac{\frac{1}{2}}{1 - ( \frac{1}{2} ) ^{2}}

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= \frac{π}{6} + \frac{\frac{1}{2}}{1 - ( \frac{1}{2} ) ^{2}}

Simplify

\frac{π}{6} + \frac{\frac{1}{2}}{1 - ( \frac{1}{2} ) ^{2}} : \frac{π}{6} + \frac{3}{3}

\frac{π}{6} + \frac{\frac{1}{2}}{1 - ( \frac{1}{2} ) ^{2}}

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

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

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

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

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

(\frac{1}{2})^{2} = \frac{1}{4}

(\frac{1}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{2 ^{2}}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{2}}

2^{2} = 4

= \frac{1}{4}

= 1 - \frac{1}{4}

Join

1 - \frac{1}{4} : \frac{3}{4}

1 - \frac{1}{4}

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{1}{4}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 4 - 1}{4}

1 \cdot 4 - 1 = 3

1 \cdot 4 - 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 - 1

Subtract the numbers:

4 - 1 = 3

= 3

= \frac{3}{4}

= \frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= \frac{1}{2 \cdot \frac{3}{2}}

Multiply

2 \cdot \frac{3}{2} : 3

2 \cdot \frac{3}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 2}{2}

Cancel the common factor:

2

= 3

= \frac{1}{3}

= \frac{π}{6} + \frac{1}{3}

\frac{1}{3} = \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{π}{6} + \frac{3}{3}

= \frac{π}{6} + \frac{3}{3}

cot (6 0^{\circ}) tan (3 0^{\circ}) + sec^{2} (4 5^{\circ}) sin (3 0^{\circ})

(1)^{2} \cdot sin (\frac{π}{3}) \cdot cos (\frac{π}{3})

\frac{π}{2} [sin (0) + sin (π)]

cos (arctan (\frac{5}{4}))

csc (- 39 0^{\circ})

What is the value of arcsin(1/2)+(1/2)/(sqrt(1-(1/2)^2)) ?
The value of arcsin(1/2)+(1/2)/(sqrt(1-(1/2)^2)) is pi/6+(sqrt(3))/3