Is cot(pi-θ)=-cot(θ) ?

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

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Popular Trigonometry >

prove cot(pi-θ)=-cot(θ)

Solution

prove

cot (π - θ) = - cot (θ)

Solution

True

Solution steps

cot (π - θ) = - cot (θ)

Manipulating left side

cot (π - θ)

Rewrite using trig identities

cot (π - θ)

Use the basic trigonometric identity:

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

= \frac{cos ( π - θ )}{sin ( π - θ )}

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

= - \frac{cos ( θ )}{sin ( θ )}

= - \frac{cos ( θ )}{sin ( θ )}

= - \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= - cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Is cot(pi-θ)=-cot(θ) ?
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