What is the general solution for tan(pi-θ)=tan(θ) ?

The general solution for tan(pi-θ)=tan(θ) is θ=2pin,θ=pi+2pin

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tan(pi-θ)=tan(θ)

Solution

tan (π - θ) = tan (θ)

Solution

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (π - θ) = tan (θ)

Rewrite using trig identities

tan (π - θ) = tan (θ)

Rewrite using trig identities

tan (π - θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( π - θ )}{cos ( π - θ )}

Use the Angle Difference identity:

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

= \frac{sin ( π ) cos ( θ ) - cos ( π ) sin ( θ )}{cos ( π - θ )}

Use the Angle Difference identity:

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

= \frac{sin ( π ) cos ( θ ) - cos ( π ) sin ( θ )}{cos ( π ) cos ( θ ) + sin ( π ) sin ( θ )}

Simplify

\frac{sin ( π ) cos ( θ ) - cos ( π ) sin ( θ )}{cos ( π ) cos ( θ ) + sin ( π ) sin ( θ )} : - \frac{sin ( θ )}{cos ( θ )}

\frac{sin ( π ) cos ( θ ) - cos ( π ) sin ( θ )}{cos ( π ) cos ( θ ) + sin ( π ) sin ( θ )}

sin (π) cos (θ) - cos (π) sin (θ) = sin (θ)

sin (π) cos (θ) - cos (π) sin (θ)

sin (π) cos (θ) = 0

sin (π) cos (θ)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot cos (θ)

Apply rule

0 \cdot a = 0

= 0

cos (π) sin (θ) = - sin (θ)

cos (π) sin (θ)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= - sin (θ)

= 0 - (- sin (θ))

Refine

= sin (θ)

= \frac{sin ( θ )}{cos ( π ) cos ( θ ) + sin ( π ) sin ( θ )}

cos (π) cos (θ) + sin (π) sin (θ) = - cos (θ)

cos (π) cos (θ) + sin (π) sin (θ)

cos (π) cos (θ) = - cos (θ)

cos (π) cos (θ)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= - cos (θ)

= - cos (θ) + sin (π) sin (θ)

sin (π) sin (θ) = 0

sin (π) sin (θ)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot sin (θ)

Apply rule

0 \cdot a = 0

= 0

= - cos (θ) + 0

- cos (θ) + 0 = - cos (θ)

= - cos (θ)

= \frac{sin ( θ )}{- cos ( θ )}

Apply the fraction rule:

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

= - \frac{sin ( θ )}{cos ( θ )}

= - \frac{sin ( θ )}{cos ( θ )}

- \frac{sin ( θ )}{cos ( θ )} = tan (θ)

- \frac{sin ( θ )}{cos ( θ )} = tan (θ)

Subtract

tan (θ)

from both sides

- \frac{sin ( θ )}{cos ( θ )} - tan (θ) = 0

Simplify

- \frac{sin ( θ )}{cos ( θ )} - tan (θ) : \frac{- sin ( θ ) - tan ( θ ) cos ( θ )}{cos ( θ )}

- \frac{sin ( θ )}{cos ( θ )} - tan (θ)

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) cos ( θ )}{cos ( θ )}

= - \frac{sin ( θ )}{cos ( θ )} - \frac{tan ( θ ) cos ( θ )}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( θ ) - tan ( θ ) cos ( θ )}{cos ( θ )}

\frac{- sin ( θ ) - tan ( θ ) cos ( θ )}{cos ( θ )} = 0

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

- sin (θ) - tan (θ) cos (θ) = 0

Rewrite using trig identities

- sin (θ) - cos (θ) tan (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - sin (θ) - cos (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

- sin (θ) - cos (θ) \frac{sin ( θ )}{cos ( θ )} : - 2 sin (θ)

- sin (θ) - cos (θ) \frac{sin ( θ )}{cos ( θ )}

cos (θ) \frac{sin ( θ )}{cos ( θ )} = sin (θ)

cos (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ)

= - sin (θ) - sin (θ)

Add similar elements:

- sin (θ) - sin (θ) = - 2 sin (θ)

= - 2 sin (θ)

= - 2 sin (θ)

- 2 sin (θ) = 0

Divide both sides by

- 2

- 2 sin (θ) = 0

Divide both sides by

- 2

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

Simplify

sin (θ) = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Popular Examples

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2 sin^{2} (t) - 3 cos (t) = 3

cos(x)=(sqrt(3))/2 ,0<= x<= 2pi

cos (x) = \frac{3}{2}, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for tan(pi-θ)=tan(θ) ?
The general solution for tan(pi-θ)=tan(θ) is θ=2pin,θ=pi+2pin