Is tan((41pi)/2+x)+cot(pi+x)=0 ?

The answer to whether tan((41pi)/2+x)+cot(pi+x)=0 is True

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prove tan((41pi)/2+x)+cot(pi+x)=0

Solution

prove

tan (\frac{41 π}{2} + x) + cot (π + x) = 0

Solution

True

Solution steps

tan (\frac{41 π}{2} + x) + cot (π + x) = 0

Manipulating left side

tan (\frac{41 π}{2} + x) + cot (π + x)

Rewrite using trig identities

tan (\frac{41 π}{2} + x)

Use the basic trigonometric identity:

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

= \frac{sin ( \frac{41 π}{2} + x )}{cos ( \frac{41 π}{2} + x )}

Use the Angle Sum identity:

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

= \frac{sin ( \frac{41 π}{2} ) cos ( x ) + cos ( \frac{41 π}{2} ) sin ( x )}{cos ( \frac{41 π}{2} + x )}

Use the Angle Sum identity:

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

= \frac{sin ( \frac{41 π}{2} ) cos ( x ) + cos ( \frac{41 π}{2} ) sin ( x )}{cos ( \frac{41 π}{2} ) cos ( x ) - sin ( \frac{41 π}{2} ) sin ( x )}

Simplify

\frac{sin ( \frac{41 π}{2} ) cos ( x ) + cos ( \frac{41 π}{2} ) sin ( x )}{cos ( \frac{41 π}{2} ) cos ( x ) - sin ( \frac{41 π}{2} ) sin ( x )} : - \frac{cos ( x )}{sin ( x )}

\frac{sin ( \frac{41 π}{2} ) cos ( x ) + cos ( \frac{41 π}{2} ) sin ( x )}{cos ( \frac{41 π}{2} ) cos ( x ) - sin ( \frac{41 π}{2} ) sin ( x )}

sin (\frac{41 π}{2}) cos (x) + cos (\frac{41 π}{2}) sin (x) = cos (x)

sin (\frac{41 π}{2}) cos (x) + cos (\frac{41 π}{2}) sin (x)

sin (\frac{41 π}{2}) cos (x) = cos (x)

sin (\frac{41 π}{2}) cos (x)

sin (\frac{41 π}{2}) = 1

sin (\frac{41 π}{2})

sin (\frac{41 π}{2}) = sin (\frac{π}{2})

sin (\frac{41 π}{2})

Rewrite

\frac{41 π}{2}

2 π \cdot 10 + \frac{π}{2}

= sin (2 π 10 + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 10 + \frac{π}{2}) = sin (\frac{π}{2})

= sin (\frac{π}{2})

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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}

= 1

= 1

= 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

cos (\frac{41 π}{2}) sin (x) = 0

cos (\frac{41 π}{2}) sin (x)

cos (\frac{41 π}{2}) = 0

cos (\frac{41 π}{2})

cos (\frac{41 π}{2}) = cos (\frac{π}{2})

cos (\frac{41 π}{2})

Rewrite

\frac{41 π}{2}

2 π \cdot 10 + \frac{π}{2}

= cos (2 π 10 + \frac{π}{2})

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 10 + \frac{π}{2}) = cos (\frac{π}{2})

= cos (\frac{π}{2})

= cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) + 0

cos (x) + 0 = cos (x)

= cos (x)

= \frac{cos ( x )}{cos ( \frac{41 π}{2} ) cos ( x ) - sin ( \frac{41 π}{2} ) sin ( x )}

cos (\frac{41 π}{2}) cos (x) - sin (\frac{41 π}{2}) sin (x) = - sin (x)

cos (\frac{41 π}{2}) cos (x) - sin (\frac{41 π}{2}) sin (x)

cos (\frac{41 π}{2}) cos (x) = 0

cos (\frac{41 π}{2}) cos (x)

cos (\frac{41 π}{2}) = 0

cos (\frac{41 π}{2})

cos (\frac{41 π}{2}) = cos (\frac{π}{2})

cos (\frac{41 π}{2})

Rewrite

\frac{41 π}{2}

2 π \cdot 10 + \frac{π}{2}

= cos (2 π 10 + \frac{π}{2})

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 10 + \frac{π}{2}) = cos (\frac{π}{2})

= cos (\frac{π}{2})

= cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{41 π}{2}) sin (x) = sin (x)

sin (\frac{41 π}{2}) sin (x)

sin (\frac{41 π}{2}) = 1

sin (\frac{41 π}{2})

sin (\frac{41 π}{2}) = sin (\frac{π}{2})

sin (\frac{41 π}{2})

Rewrite

\frac{41 π}{2}

2 π \cdot 10 + \frac{π}{2}

= sin (2 π 10 + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 10 + \frac{π}{2}) = sin (\frac{π}{2})

= sin (\frac{π}{2})

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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}

= 1

= 1

= 1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

= 0 - sin (x)

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

= - sin (x)

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

cot (π + x)

Use the basic trigonometric identity:

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

= \frac{cos ( π + x )}{sin ( π + x )}

Use the Angle Sum identity:

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

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

Use the Angle Sum identity:

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

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

Simplify

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

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

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

cos (π) cos (x) - sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

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 (x)

Multiply:

1 \cdot cos (x) = cos (x)

= - cos (x)

= - cos (x) - sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

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

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

sin (π) cos (x) + cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

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

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

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 (x)

Multiply:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - sin (x)

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

= - sin (x)

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

Apply the fraction rule:

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

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

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

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

Add similar elements:

- \frac{cos ( x )}{sin ( x )} + \frac{cos ( x )}{sin ( x )} = 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan((41pi)/2+x)+cot(pi+x)=0 ?
The answer to whether tan((41pi)/2+x)+cot(pi+x)=0 is True