Is sec(pi-x)=-sec(x) ?

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

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prove sec(pi-x)=-sec(x)

Solution

prove

sec (π - x) = - sec (x)

Solution

True

Solution steps

sec (π - x) = - sec (x)

Manipulating left side

sec (π - x)

Rewrite using trig identities

sec (π - x)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( π - x )}

Use the Angle Difference identity:

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

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

Simplify

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

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

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

cos (π) cos (x) + sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

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 \cdot cos (x)

Multiply:

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

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

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 (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= - sec (x)

- sec (x)

- sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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