What is the general solution for (tan(2θ))/2 =tan(θ) ?

The general solution for (tan(2θ))/2 =tan(θ) is θ=pin

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(tan(2θ))/2 =tan(θ)

Solution

\frac{tan ( 2 θ )}{2} = tan (θ)

Solution

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

\frac{tan ( 2 θ )}{2} = tan (θ)

Subtract

tan (θ)

from both sides

\frac{tan ( 2 θ )}{2} - tan (θ) = 0

Simplify

\frac{tan ( 2 θ )}{2} - tan (θ) : \frac{tan ( 2 θ ) - 2 tan ( θ )}{2}

\frac{tan ( 2 θ )}{2} - tan (θ)

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) 2}{2}

= \frac{tan ( 2 θ )}{2} - \frac{tan ( θ ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( 2 θ ) - tan ( θ ) \cdot 2}{2}

\frac{tan ( 2 θ ) - 2 tan ( θ )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

tan (2 θ) - 2 tan (θ) = 0

Rewrite using trig identities

tan (2 θ) - 2 tan (θ)

Use the Double Angle identity:

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

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

Simplify

\frac{2 tan ( θ )}{1 - tan ^{2} ( θ )} - 2 tan (θ) : \frac{2 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

2 tan (θ) - 2 tan (θ) (1 - tan^{2} (θ)) : 2 tan^{3} (θ)

2 tan (θ) - 2 tan (θ) (1 - tan^{2} (θ))

Expand

- 2 tan (θ) (1 - tan^{2} (θ)) : - 2 tan (θ) + 2 tan^{3} (θ)

- 2 tan (θ) (1 - tan^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2 tan (θ), b = 1, c = tan^{2} (θ)

= - 2 tan (θ) \cdot 1 - (- 2 tan (θ)) tan^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 \cdot tan (θ) + 2 tan^{2} (θ) tan (θ)

Simplify

- 2 \cdot 1 \cdot tan (θ) + 2 tan^{2} (θ) tan (θ) : - 2 tan (θ) + 2 tan^{3} (θ)

- 2 \cdot 1 \cdot tan (θ) + 2 tan^{2} (θ) tan (θ)

2 \cdot 1 \cdot tan (θ) = 2 tan (θ)

2 \cdot 1 \cdot tan (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 tan (θ)

2 tan^{2} (θ) tan (θ) = 2 tan^{3} (θ)

2 tan^{2} (θ) tan (θ)

Apply exponent rule:

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

tan^{2} (θ) tan (θ) = tan^{2 + 1} (θ)

= 2 tan^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 tan^{3} (θ)

= - 2 tan (θ) + 2 tan^{3} (θ)

= - 2 tan (θ) + 2 tan^{3} (θ)

= 2 tan (θ) - 2 tan (θ) + 2 tan^{3} (θ)

Add similar elements:

2 tan (θ) - 2 tan (θ) = 0

= 2 tan^{3} (θ)

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

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

\frac{2 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )} = 0

Solve by substitution

\frac{2 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )} = 0

Let:

tan (θ) = u

\frac{2 u ^{3}}{1 - u ^{2}} = 0

\frac{2 u ^{3}}{1 - u ^{2}} = 0 : u = 0

\frac{2 u ^{3}}{1 - u ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 u^{3} = 0

Solve

2 u^{3} = 0 : u = 0

2 u^{3} = 0

Divide both sides by

2

2 u^{3} = 0

Divide both sides by

2

2 u^{3} = 0

Divide both sides by

2

\frac{2 u ^{3}}{2} = \frac{0}{2}

Simplify

u^{3} = 0

u^{3} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

and compare to zero

Solve

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

1 - u^{2} - 1 = 0 - 1

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

u = 0

Substitute back

u = tan (θ)

tan (θ) = 0

tan (θ) = 0

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

Combine all the solutions

θ = πn

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(2θ))/2 =tan(θ) ?
The general solution for (tan(2θ))/2 =tan(θ) is θ=pin