Is tan(2pi-θ)=-tan(θ) ?

The answer to whether tan(2pi-θ)=-tan(θ) is True

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Popular Trigonometry >

prove tan(2pi-θ)=-tan(θ)

Solution

prove

tan (2 π - θ) = - tan (θ)

Solution

True

Solution steps

tan (2 π - θ) = - tan (θ)

Manipulating left side

tan (2 π - θ)

Rewrite using trig identities

tan (2 π - θ)

Use the basic trigonometric identity:

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

= \frac{sin ( 2 π - θ )}{cos ( 2 π - θ )}

Use the Angle Difference identity:

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

= \frac{sin ( 2 π ) cos ( θ ) - cos ( 2 π ) sin ( θ )}{cos ( 2 π - θ )}

Use the Angle Difference identity:

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

= \frac{sin ( 2 π ) cos ( θ ) - cos ( 2 π ) sin ( θ )}{cos ( 2 π ) cos ( θ ) + sin ( 2 π ) sin ( θ )}

Simplify

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

\frac{sin ( 2 π ) cos ( θ ) - cos ( 2 π ) sin ( θ )}{cos ( 2 π ) cos ( θ ) + sin ( 2 π ) sin ( θ )}

sin (2 π) cos (θ) - cos (2 π) sin (θ) = - sin (θ)

sin (2 π) cos (θ) - cos (2 π) sin (θ)

sin (2 π) cos (θ) = 0

sin (2 π) cos (θ)

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (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

= 0 \cdot cos (θ)

Apply rule

0 \cdot a = 0

= 0

cos (2 π) sin (θ) = sin (θ)

cos (2 π) sin (θ)

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

Rewrite

2 π

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= 1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

= 0 - sin (θ)

0 - sin (θ) = - sin (θ)

= - sin (θ)

= \frac{- sin ( θ )}{cos ( 2 π ) cos ( θ ) + sin ( 2 π ) sin ( θ )}

cos (2 π) cos (θ) + sin (2 π) sin (θ) = cos (θ)

cos (2 π) cos (θ) + sin (2 π) sin (θ)

cos (2 π) cos (θ) = cos (θ)

cos (2 π) cos (θ)

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

Rewrite

2 π

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= 1 \cdot cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

sin (2 π) sin (θ) = 0

sin (2 π) sin (θ)

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (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

= 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 )}

Use the basic trigonometric identity:

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

= - tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Is tan(2pi-θ)=-tan(θ) ?
The answer to whether tan(2pi-θ)=-tan(θ) is True