Is tan(θ)=tan(pi+θ) ?

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

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Popular Trigonometry >

prove tan(θ)=tan(pi+θ)

Solution

prove

tan (θ) = tan (π + θ)

Solution

True

Solution steps

tan (θ) = tan (π + θ)

Manipulating right side

tan (π + 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 Sum 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 Sum 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)

0 - sin (θ) = - sin (θ)

= - 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 )}

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(θ)=tan(pi+θ) ?
The answer to whether tan(θ)=tan(pi+θ) is True