Is sec(θ-pi)=-sec(θ) ?

The answer to whether sec(θ-pi)=-sec(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sec(θ-pi)=-sec(θ)

Solution

prove

sec (θ - π) = - sec (θ)

Solution

True

Solution steps

sec (θ - π) = - sec (θ)

Manipulating left side

sec (θ - π)

Rewrite using trig identities

sec (θ - π)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ - π )}

Use the Angle Difference identity:

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

= \frac{1}{cos ( θ ) cos ( π ) + sin ( θ ) sin ( π )}

Simplify

\frac{1}{cos ( θ ) cos ( π ) + sin ( θ ) sin ( π )} : - \frac{1}{cos ( θ )}

\frac{1}{cos ( θ ) cos ( π ) + sin ( θ ) sin ( π )}

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

cos (θ) cos (π) + sin (θ) sin (π)

cos (θ) cos (π) = - cos (θ)

cos (θ) cos (π)

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) cos (θ)

Refine

= - cos (θ)

= - cos (θ) + sin (π) sin (θ)

sin (θ) sin (π) = 0

sin (θ) sin (π)

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 (θ)

Apply rule

0 \cdot a = 0

= 0

= - cos (θ) + 0

- cos (θ) + 0 = - cos (θ)

= - cos (θ)

= \frac{1}{- cos ( θ )}

Apply the fraction rule:

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

= - \frac{1}{cos ( θ )}

= - \frac{1}{cos ( θ )}

= - \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

- \frac{1}{\frac{1}{s e c ( θ )}}

Simplify

- \frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

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

= - \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= - sec (θ)

- sec (θ)

- sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove cos(0+pi/2)=-sin(θ)prove

cos (0 + \frac{π}{2}) = - sin (θ)

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

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

prove cos(2*x)=2*cos^2(x)-1 prove

cos (2 \cdot x) = 2 \cdot cos^{2} (x) - 1

prove csc(x)tan(x)+sec(x)=2sec(x)prove

csc (x) tan (x) + sec (x) = 2 sec (x)

prove csc(x)(sin(x)+tan(x))=1+sec(x)prove

csc (x) (sin (x) + tan (x)) = 1 + sec (x)

Frequently Asked Questions (FAQ)

Is sec(θ-pi)=-sec(θ) ?
The answer to whether sec(θ-pi)=-sec(θ) is True