What is the value of (2tan((5pi)/8))/(1-tan^2((5pi)/8)) ?

The value of (2tan((5pi)/8))/(1-tan^2((5pi)/8)) is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2tan((5pi)/8))/(1-tan^2((5pi)/8))

\frac{2 tan ( \frac{5 π}{8} )}{1 - tan ^{2} ( \frac{5 π}{8} )}

1

Solution steps

\frac{2 tan ( \frac{5 π}{8} )}{1 - tan ^{2} ( \frac{5 π}{8} )}

Rewrite using trig identities:

tan (\frac{π}{4})

\frac{2 tan ( \frac{5 π}{8} )}{1 - tan ^{2} ( \frac{5 π}{8} )}

Use the Double Angle identity:

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

= tan (2 \cdot \frac{5 π}{8})

Simplify:

2 \cdot \frac{5 π}{8} = \frac{5 π}{4}

2 \cdot \frac{5 π}{8}

Multiply fractions:

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

= \frac{5 π 2}{8}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{8}

Cancel the common factor:

2

= \frac{5 π}{4}

= tan (\frac{5 π}{4})

tan (\frac{5 π}{4}) = tan (\frac{π}{4})

tan (\frac{5 π}{4})

Rewrite

\frac{5 π}{4}

π + \frac{π}{4}

= tan (π + \frac{π}{4})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{π}{4}) = tan (\frac{π}{4})

= tan (\frac{π}{4})

= tan (\frac{π}{4})

= tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

800 \cdot cos (3 0^{\circ})

3 sec^{2} (\frac{3 π}{4})

2 sin (\frac{2 π}{5}) cos (\frac{2 π}{5})

\frac{sin ( 84 0 ^{\circ} )}{cos ( 33 0 ^{\circ} ) + sin ( 31 5 ^{\circ} )}

cos (arctan (3) + arcsin (\frac{1}{3}))

What is the value of (2tan((5pi)/8))/(1-tan^2((5pi)/8)) ?
The value of (2tan((5pi)/8))/(1-tan^2((5pi)/8)) is 1