How do you find the value for ${tan}^{- 1} [tan (5 \frac{π}{7})]$ $tan^-1[tan(5pi/7)]$ ?

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How do you find the value for ${tan}^{- 1} [tan (5 \frac{π}{7})]$ $tan^-1[tan(5pi/7)]$ ?

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Answer 1 · 2015-05-28T22:17:35

The formula
${tan}^{- 1} [tan (a)] = a$ $tan^(-1)[tan(a)]=a$
works for $a \in (- \frac{π}{2}; \frac{π}{2})$ $a in (-pi/2;pi/2)$

In our case $\frac{5 π}{7} > \frac{π}{2}$ $(5pi)/7>pi/2$ but we can use the periodicity of $tan$ $tan$ :
$tan (\frac{5 π}{7}) = tan (\frac{5 π}{7} - π) = tan (- \frac{2 π}{7})$ $tan((5pi)/7)=tan((5pi)/7-pi)=tan(-(2pi)/7)$
${tan}^{- 1} [tan (\frac{5 π}{7})] = {tan}^{- 1} [tan (- \frac{2 π}{7})] = - \frac{2 π}{7}$ $tan^(-1)[tan((5pi)/7)]=tan^(-1)[tan(-(2pi)/7)]=-(2pi)/7$

Btw, a similar formula
$tan [{tan}^{- 1} (a)] = a$ $tan[tan^(-1)(a)]=a$
works for all $a in RR$ .

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How do you find the value for ${tan}^{- 1} [tan (5 \frac{π}{7})]$ $tan^-1[tan(5pi/7)]$ ?

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How do you find the value for ${tan}^{- 1} [tan (5 \frac{π}{7})]$ $tan^-1[tan(5pi/7)]$ ?