Is cot(x-pi/2)=-tan(x) ?

The answer to whether cot(x-pi/2)=-tan(x) is True

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prove cot(x-pi/2)=-tan(x)

Solution

prove

cot (x - \frac{π}{2}) = - tan (x)

Solution

True

Solution steps

cot (x - \frac{π}{2}) = - tan (x)

Manipulating left side

cot (x - \frac{π}{2})

Rewrite using trig identities

cot (x - \frac{π}{2})

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

cos (x) cos (\frac{π}{2}) = 0

cos (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (x) sin (\frac{π}{2}) = sin (x)

sin (x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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 \cdot sin (x)

Multiply:

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

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

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

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

sin (x) cos (\frac{π}{2}) = 0

sin (x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 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}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

cos (x) sin (\frac{π}{2}) = cos (x)

cos (x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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 \cdot cos (x)

Multiply:

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

= cos (x)

= 0 - cos (x)

0 - cos (x) = - 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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Frequently Asked Questions (FAQ)

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