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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (\frac{π}{2} - x) cot (x)

Rewrite using trig identities

cos (\frac{π}{2} - x)

Use the Angle Difference identity:

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

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

Simplify

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)

= sin (x)

= sin (x) cot (x)

Express with sin, cos

cot (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( x )}{sin ^{2} ( x )} = cot (x) csc (x)

2 sin (9 π - x) = 2 sin (x)

1 csc (θ) - sin (θ) = cot (θ) cos (θ)

\frac{sec ( x )}{csc ^{2} ( x )} = sec (x) - cos (x)

\frac{1}{sin ( θ ) + 1} = \frac{1 - sin ( θ )}{cos ^{2} ( θ )}

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