Is tan(-sqrt(7))=tan(pi-sqrt(7)) ?

The answer to whether tan(-sqrt(7))=tan(pi-sqrt(7)) is True

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prove tan(-sqrt(7))=tan(pi-sqrt(7))

Solution

prove

tan (- 7) = tan (π - 7)

Solution

True

Solution steps

tan (- 7) = tan (π - 7)

Manipulating left side

tan (- 7)

Use the negative angle identity:

tan (- x) = - tan (x)

= - tan (7)

Express with sin, cos

- tan (7)

Use the basic trigonometric identity:

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

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

Manipulating right side

tan (π - 7)

Rewrite using trig identities

tan (π - 7)

Use the basic trigonometric identity:

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

= \frac{sin ( π - 7 )}{cos ( π - 7 )}

Use the Angle Difference identity:

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

= \frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π - 7 )}

Use the Angle Difference identity:

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

= \frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

\frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )} = - \frac{sin ( 7 )}{cos ( 7 )}

\frac{sin ( π ) cos ( 7 ) - cos ( π ) sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

sin (π) cos (7) - cos (π) sin (7) = sin (7)

sin (π) cos (7) - cos (π) sin (7)

sin (π) cos (7) = 0

sin (π) cos (7)

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 cos (7)

Apply rule

0 \cdot a = 0

= 0

cos (π) sin (7) = - sin (7)

cos (π) sin (7)

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 sin (7)

Multiply:

1 \cdot sin (7) = sin (7)

= - sin (7)

= 0 - (- sin (7))

Apply rule

- (- a) = a

= 0 + sin (7)

0 + sin (7) = sin (7)

= sin (7)

= \frac{sin ( 7 )}{cos ( π ) cos ( 7 ) + sin ( π ) sin ( 7 )}

cos (π) cos (7) + sin (π) sin (7) = - cos (7)

cos (π) cos (7) + sin (π) sin (7)

cos (π) cos (7) = - cos (7)

cos (π) cos (7)

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 (7)

Multiply:

1 \cdot cos (7) = cos (7)

= - cos (7)

= - cos (7) + sin (π) sin (7)

sin (π) sin (7) = 0

sin (π) sin (7)

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 (7)

Apply rule

0 \cdot a = 0

= 0

= - cos (7) + 0

- cos (7) + 0 = - cos (7)

= - cos (7)

= \frac{sin ( 7 )}{- cos ( 7 )}

Apply the fraction rule:

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

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

= - \frac{sin ( 7 )}{cos ( 7 )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan(-sqrt(7))=tan(pi-sqrt(7)) ?
The answer to whether tan(-sqrt(7))=tan(pi-sqrt(7)) is True