Is tan(x-pi)=tan(x) ?

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

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Popular Trigonometry >

prove tan(x-pi)=tan(x)

Solution

prove

tan (x - π) = tan (x)

Solution

True

Solution steps

tan (x - π) = tan (x)

Manipulating left side

tan (x - π)

Rewrite using trig identities

tan (x - π)

Use the basic trigonometric identity:

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

= \frac{sin ( x - π )}{cos ( x - π )}

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

sin (x) cos (π) - cos (x) sin (π) = - sin (x)

sin (x) cos (π) - cos (x) sin (π)

sin (x) cos (π) = - sin (x)

sin (x) cos (π)

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) sin (x)

Refine

= - sin (x)

= - sin (x) - sin (π) cos (x)

cos (x) sin (π) = 0

cos (x) sin (π)

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 (x)

Apply rule

0 \cdot a = 0

= 0

= - sin (x) - 0

- sin (x) - 0 = - sin (x)

= - sin (x)

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

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

cos (x) cos (π) + sin (x) sin (π)

cos (x) cos (π) = - cos (x)

cos (x) cos (π)

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) cos (x)

Refine

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (x) sin (π) = 0

sin (x) sin (π)

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 (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

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

Apply the fraction rule:

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

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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