What is the general solution for arccos(x)-arcsin(x)=arcsin(1-x) ?

The general solution for arccos(x)-arcsin(x)=arcsin(1-x) is x=0,x= 1/2

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arccos(x)-arcsin(x)=arcsin(1-x)

Solution

arccos (x) - arcsin (x) = arcsin (1 - x)

Solution

x = 0, x = \frac{1}{2}

Solution steps

arccos (x) - arcsin (x) = arcsin (1 - x)

a = b \Rightarrow sin (a) = sin (b)

sin (arccos (x) - arcsin (x)) = sin (arcsin (1 - x))

Use the following identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

sin (arccos (x)) cos (arcsin (x)) - cos (arccos (x)) sin (arcsin (x)) = sin (arcsin (1 - x))

Use the following identity:

sin (arccos (x)) = 1 - x^{2}

Use the following identity:

cos (arcsin (x)) = 1 - x^{2}

Use the following identity:

cos (arccos (x)) = x

Use the following identity:

sin (arcsin (x)) = x

1 - x^{2} 1 - x^{2} - xx = 1 - x

Solve

1 - x^{2} 1 - x^{2} - xx = 1 - x : x = 0, x = \frac{1}{2}

1 - x^{2} 1 - x^{2} - xx = 1 - x

Expand

1 - x^{2} 1 - x^{2} - xx : 1 - 2 x^{2}

1 - x^{2} 1 - x^{2} - xx

1 - x^{2} 1 - x^{2} = 1 - x^{2}

1 - x^{2} 1 - x^{2}

Apply radical rule:

a a = a

1 - x^{2} 1 - x^{2} = 1 - x^{2}

= 1 - x^{2}

xx = x^{2}

xx

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= x^{1 + 1}

Add the numbers:

1 + 1 = 2

= x^{2}

= 1 - x^{2} - x^{2}

Refine

= 1 - 2 x^{2}

1 - 2 x^{2} = 1 - x

Solve

1 - 2 x^{2} = 1 - x : x = 0, x = \frac{1}{2}

1 - 2 x^{2} = 1 - x

Move

x

to the left side

1 - 2 x^{2} = 1 - x

Add

x

to both sides

1 - 2 x^{2} + x = 1 - x + x

Simplify

1 - 2 x^{2} + x = 1

1 - 2 x^{2} + x = 1

Move

1

to the left side

1 - 2 x^{2} + x = 1

Subtract

1

from both sides

1 - 2 x^{2} + x - 1 = 1 - 1

Simplify

- 2 x^{2} + x = 0

- 2 x^{2} + x = 0

Solve with the quadratic formula

- 2 x^{2} + x = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 0

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 0 = 1

1^{2} - 4 (- 2) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

x_{1, 2} = \frac{- 1 \pm 1}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 1 + 1}{2 ( - 2 )}, x_{2} = \frac{- 1 - 1}{2 ( - 2 )}

x = \frac{- 1 + 1}{2 ( - 2 )} : 0

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

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

x = \frac{- 1 - 1}{2 ( - 2 )} : \frac{1}{2}

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

Verify Solutions:

x = 0

True

, x = \frac{1}{2}

True

Check the solutions by plugging them into

1 - x^{2} 1 - x^{2} - xx = 1 - x

Remove the ones that don't agree with the equation.

Plug in

x = 0 :

True

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0 = 1 - 0

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0 = 1 - 0

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0

Apply rule

0^{a} = 0

0^{2} = 0

= 1 - 0 1 - 0 - 0 \cdot 0

1 - 0 1 - 0 = 1

1 - 0 1 - 0

Apply radical rule:

a a = a

1 - 0 1 - 0 = 1 - 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

0 \cdot 0 = 0

0 \cdot 0

Multiply the numbers:

0 \cdot 0 = 0

= 0

= 1 - 0

1 - 0 = 1 - 0

True

Plug in

x = \frac{1}{2} :

True

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

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

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

Remove parentheses:

(a) = a

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

1 - (\frac{1}{2})^{2} 1 - (\frac{1}{2})^{2} = \frac{3}{4}

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

Apply radical rule:

a a = a

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

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{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

\frac{1}{2} \cdot \frac{1}{2}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

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

Apply rule

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

= \frac{3 - 1}{4}

Subtract the numbers:

3 - 1 = 2

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Remove parentheses:

(a) = a

= 1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

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

True

The solutions are

x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arccos (x) - arcsin (x) = arcsin (1 - x)

Remove the ones that don't agree with the equation.

Check the solution

0 :

True

0

Plug in

n = 1

0

For

arccos (x) - arcsin (x) = arcsin (1 - x)

plug in

x = 0

arccos (0) - arcsin (0) = arcsin (1 - 0)

Refine

1.57079 \dots = 1.57079 \dots

\Rightarrow True

Check the solution

\frac{1}{2} :

True

\frac{1}{2}

Plug in

n = 1

\frac{1}{2}

For

arccos (x) - arcsin (x) = arcsin (1 - x)

plug in

x = \frac{1}{2}

arccos (\frac{1}{2}) - arcsin (\frac{1}{2}) = arcsin (1 - \frac{1}{2})

Refine

0.52359 \dots = 0.52359 \dots

\Rightarrow True

x = 0, x = \frac{1}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for arccos(x)-arcsin(x)=arcsin(1-x) ?
The general solution for arccos(x)-arcsin(x)=arcsin(1-x) is x=0,x= 1/2