What is the value of tan((3pi)/4) ?

The value of tan((3pi)/4) is -1

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tan((3pi)/4)

tan (\frac{3 π}{4})

- 1

Solution steps

tan (\frac{3 π}{4})

Rewrite using trig identities:

\frac{sin ( \frac{3 π}{4} )}{cos ( \frac{3 π}{4} )}

tan (\frac{3 π}{4})

Use the basic trigonometric identity:

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

= \frac{sin ( \frac{3 π}{4} )}{cos ( \frac{3 π}{4} )}

= \frac{sin ( \frac{3 π}{4} )}{cos ( \frac{3 π}{4} )}

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 π}{4})

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}

= \frac{2}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

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}

= - \frac{2}{2}

= \frac{\frac{2}{2}}{- \frac{2}{2}}

Simplify

\frac{\frac{2}{2}}{- \frac{2}{2}} : - 1

\frac{\frac{2}{2}}{- \frac{2}{2}}

Apply the fraction rule:

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

= - \frac{\frac{2}{2}}{\frac{2}{2}}

Divide fractions:

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

= - \frac{2 \cdot 2}{2 2}

Cancel the common factor:

2

= - \frac{2}{2}

Cancel the common factor:

2

= - 1

= - 1

sin^{2} (x) + cos^{2} (x) = 1

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

arccsc (2)

tan (5)

sin^{3} (0)