Is sin(pi/2-x)cot(pi/2+x)=-sin(x) ?

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

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prove sin(pi/2-x)cot(pi/2+x)=-sin(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

sin (\frac{π}{2} - x)

Use the Angle Difference identity:

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

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

Simplify

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

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

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

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:

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

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

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) - 0

cos (x) - 0 = cos (x)

= cos (x)

= cos (x)

= cos (x) cot (\frac{π}{2} + x)

Rewrite using trig identities

cot (\frac{π}{2} + x)

Use the basic trigonometric identity:

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

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

Use the Angle Sum identity:

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

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

Use the Angle Sum identity:

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

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

Simplify

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

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

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

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

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (x)

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 (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (x)

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:

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

= sin (x)

= 0 - sin (x)

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

= - sin (x)

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

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

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

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

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:

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

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

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) + 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 )}

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

cos (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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