What is the general solution for arctan(x)+arctan(1-x)=arctan(4/3) ?

The general solution for arctan(x)+arctan(1-x)=arctan(4/3) is x= 1/2

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arctan(x)+arctan(1-x)=arctan(4/3)

Solution

arctan (x) + arctan (1 - x) = arctan (\frac{4}{3})

Solution

x = \frac{1}{2}

Solution steps

arctan (x) + arctan (1 - x) = arctan (\frac{4}{3})

Rewrite using trig identities

arctan (x) + arctan (1 - x)

Use the Sum to Product identity:

arctan (s) + arctan (t) = arctan (\frac{s + t}{1 - s t})

= arctan (\frac{x + 1 - x}{1 - x ( 1 - x )})

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{4}{3})

Apply trig inverse properties

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{4}{3})

arctan (x) = a \Rightarrow x = tan (a)

\frac{x + 1 - x}{1 - x ( 1 - x )} = tan (arctan (\frac{4}{3}))

tan (arctan (\frac{4}{3})) = \frac{4}{3}

tan (arctan (\frac{4}{3}))

Rewrite using trig identities:

tan (arctan (\frac{4}{3})) = \frac{4}{3}

Use the following identity:

tan (arctan (x)) = x

= \frac{4}{3}

= \frac{4}{3}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{4}{3}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{4}{3}

Solve

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{4}{3} : x = \frac{1}{2}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{4}{3}

Cross multiply

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{4}{3}

Simplify

\frac{x + 1 - x}{1 - x ( 1 - x )} : \frac{1}{1 - x ( 1 - x )}

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

x + 1 - x = 1

x + 1 - x

Group like terms

= x - x + 1

Add similar elements:

x - x = 0

= 1

= \frac{1}{1 - x ( - x + 1 )}

\frac{1}{1 - x ( 1 - x )} = \frac{4}{3}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

1 \cdot 3 = (1 - x (1 - x)) \cdot 4

Simplify

1 \cdot 3 : 3

1 \cdot 3

Multiply the numbers:

1 \cdot 3 = 3

= 3

3 = (1 - x (1 - x)) \cdot 4

3 = (1 - x (1 - x)) \cdot 4

Solve

3 = (1 - x (1 - x)) \cdot 4 : x = \frac{1}{2}

3 = (1 - x (1 - x)) \cdot 4

Expand

(1 - x (1 - x)) \cdot 4 : 4 - 4 x + 4 x^{2}

(1 - x (1 - x)) \cdot 4

Expand

1 - x (1 - x) : 1 - x + x^{2}

1 - x (1 - x)

Expand

- x (1 - x) : - x + x^{2}

- x (1 - x)

Apply the distributive law:

a (b - c) = ab - a c

a = - x, b = 1, c = x

= - x \cdot 1 - (- x) x

Apply minus-plus rules

- (- a) = a

= - 1 \cdot x + xx

Simplify

- 1 \cdot x + xx : - x + x^{2}

- 1 \cdot x + xx

1 \cdot x = x

1 \cdot x

Multiply:

1 \cdot x = x

= x

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}

= - x + x^{2}

= - x + x^{2}

= 1 - x + x^{2}

= 4 (x^{2} - x + 1)

= 4 (1 - x + x^{2})

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

= 4 \cdot 1 - 4 x + 4 x^{2}

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 x + 4 x^{2}

3 = 4 - 4 x + 4 x^{2}

Switch sides

4 - 4 x + 4 x^{2} = 3

Move

3

to the left side

4 - 4 x + 4 x^{2} = 3

Subtract

3

from both sides

4 - 4 x + 4 x^{2} - 3 = 3 - 3

Simplify

4 x^{2} - 4 x + 1 = 0

4 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

4 x^{2} - 4 x + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 4, c = 1

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

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

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

x = \frac{- ( - 4 )}{2 \cdot 4}

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

\frac{- ( - 4 )}{2 \cdot 4}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

x = \frac{1}{2}

The solution to the quadratic equation is:

x = \frac{1}{2}

x = \frac{1}{2}

x = \frac{1}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (x) + arctan (1 - x) = arctan (\frac{4}{3})

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

Check the solution

\frac{1}{2} :

True

\frac{1}{2}

Plug in

n = 1

\frac{1}{2}

For

arctan (x) + arctan (1 - x) = arctan (\frac{4}{3})

plug in

x = \frac{1}{2}

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

Refine

0.92729 \dots = 0.92729 \dots

\Rightarrow True

x = \frac{1}{2}

Graph

Popular Examples

cos(θ)=(-8)/(sqrt(14)*\sqrt{20)}

cos (θ) = \frac{- 8}{14 \cdot 20}

sin^2(x)= 1/3

sin^{2} (x) = \frac{1}{3}

cot(x)= 2/3

cot (x) = \frac{2}{3}

4sin^2(x)+1=2

4 sin^{2} (x) + 1 = 2

cos(2x)=-0.8

cos (2 x) = - 0.8

Frequently Asked Questions (FAQ)

What is the general solution for arctan(x)+arctan(1-x)=arctan(4/3) ?
The general solution for arctan(x)+arctan(1-x)=arctan(4/3) is x= 1/2