Is sec((3pi)/2-x)=-csc(x) ?

The answer to whether sec((3pi)/2-x)=-csc(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sec((3pi)/2-x)=-csc(x)

Solution

prove

sec (\frac{3 π}{2} - x) = - csc (x)

Solution

True

Solution steps

sec (\frac{3 π}{2} - x) = - csc (x)

Manipulating left side

sec (\frac{3 π}{2} - x)

Rewrite using trig identities

sec (\frac{3 π}{2} - x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{cos ( \frac{3 π}{2} - x )}

Use the Angle Difference identity:

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

= \frac{1}{cos ( \frac{3 π}{2} ) cos ( x ) + sin ( \frac{3 π}{2} ) sin ( x )}

Simplify

\frac{1}{cos ( \frac{3 π}{2} ) cos ( x ) + sin ( \frac{3 π}{2} ) sin ( x )} : - \frac{1}{sin ( x )}

\frac{1}{cos ( \frac{3 π}{2} ) cos ( x ) + sin ( \frac{3 π}{2} ) sin ( x )}

cos (\frac{3 π}{2}) cos (x) + sin (\frac{3 π}{2}) sin (x) = - sin (x)

cos (\frac{3 π}{2}) cos (x) + sin (\frac{3 π}{2}) sin (x)

cos (\frac{3 π}{2}) cos (x) = 0

cos (\frac{3 π}{2}) cos (x)

cos (\frac{3 π}{2}) = 0

cos (\frac{3 π}{2})

Rewrite using trig identities:

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

cos (\frac{3 π}{2})

Write

cos (\frac{3 π}{2})

cos (π + \frac{π}{2})

= cos (π + \frac{π}{2})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

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

Use the following trivial identity:

sin (π) = 0

sin (π)

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

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) \cdot 0 - 0 \cdot 1

Simplify

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{3 π}{2}) sin (x) = - sin (x)

sin (\frac{3 π}{2}) sin (x)

sin (\frac{3 π}{2}) = - 1

sin (\frac{3 π}{2})

Rewrite using trig identities:

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

sin (\frac{3 π}{2})

Write

sin (\frac{3 π}{2})

sin (π + \frac{π}{2})

= sin (π + \frac{π}{2})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

sin (π) = 0

sin (π)

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

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

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

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

= 0 \cdot 0 + (- 1) \cdot 1

Simplify

= - 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{1}{- sin ( x )}

Apply the fraction rule:

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

= - \frac{1}{sin ( x )}

= - \frac{1}{sin ( x )}

= - \frac{1}{sin ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

- \frac{1}{\frac{1}{c s c ( x )}}

Simplify

- \frac{1}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= - \frac{csc ( x )}{1}

Apply rule

\frac{a}{1} = a

= - csc (x)

- csc (x)

- csc (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove ((sec^2(x)-1))/(sec^2(x))=sin^2(x)prove

\frac{( sec ^{2} ( x ) - 1 )}{sec ^{2} ( x )} = sin^{2} (x)

prove cos(x)+sin(x)tan(x)= 1/(cos(x))prove

cos (x) + sin (x) tan (x) = \frac{1}{cos ( x )}

prove cos(α+β)-cos(α-β)=-2sin(α)sin(β)prove

cos (α + β) - cos (α - β) = - 2 sin (α) sin (β)

prove tan(θ)*csc(θ)= 1/(cos(θ))prove

tan (θ) \cdot csc (θ) = \frac{1}{cos ( θ )}

prove (sec^2(x))/(tan^2(x))=1+cot^2(x)prove

\frac{sec ^{2} ( x )}{tan ^{2} ( x )} = 1 + cot^{2} (x)

Frequently Asked Questions (FAQ)

Is sec((3pi)/2-x)=-csc(x) ?
The answer to whether sec((3pi)/2-x)=-csc(x) is True