What is the value of arcsin(sqrt(1.5^2-1^2)) ?

The value of arcsin(sqrt(1.5^2-1^2)) is No Solution for x\in\mathbb{R}

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arcsin(sqrt(1.5^2-1^2))

arcsin (1. 5^{2} - 1^{2})

No Solution for x \in R

Solution steps

arcsin (1. 5^{2} - 1^{2})

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

Simplify:

(\frac{3}{2})^{2} - 1^{2} = \frac{5}{2}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Apply exponent rule:

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

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

Join

\frac{3 ^{2}}{2 ^{2}} - 1 : \frac{5}{4}

\frac{3 ^{2}}{2 ^{2}} - 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

3^{2} - 1 \cdot 2^{2} = 5

3^{2} - 1 \cdot 2^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

1 \cdot 2^{2} = 4

1 \cdot 2^{2}

Multiply:

1 \cdot 2^{2} = 2^{2}

= 2^{2}

2^{2} = 4

= 4

= 9 - 4

Subtract the numbers:

9 - 4 = 5

= 5

= \frac{5}{2 ^{2}}

2^{2} = 4

= \frac{5}{4}

= \frac{5}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{5}{2}

= arcsin (\frac{5}{2})

Real domain for

arcsin (x)

- 1 \leq x \leq 1

No Solution for x \in R

\frac{cos ^{5} ( 0 )}{5}

cos (0) - cos (\frac{π}{2})

\frac{40.5}{tan ( 5 )}

sec (4 5^{\circ}) csc (4 5^{\circ}) - 1

9 + cos (2) - cos (1)

What is the value of arcsin(sqrt(1.5^2-1^2)) ?
The value of arcsin(sqrt(1.5^2-1^2)) is No Solution for x\in\mathbb{R}