illusiia

2021-08-15

If $f\left(\theta \right)=\mathrm{tan}\theta$ and $f\left(a\right)=2,$ find the exact value of:
a) $f\left(-a\right)$
b) $f\left(a\right)+f\left(a+\pi \right)+f\left(a+2\pi \right)$

Bella

1) $f\left(\theta \right)=\mathrm{tan}\theta$
2) $f\left(a\right)=2$
a) To find $f\left(-a\right)$
Substitute $\left(-\theta \right)$ in place of $\theta$ in (1):
$f\left(-\theta \right)=\mathrm{tan}\left(-\theta \right)$
From the tangent trigonometric function's characteristic:
$\mathrm{tan}\left(-\theta \right)=-\mathrm{tan}\theta$
Thus,
3) $f\left(-\theta \right)=-\mathrm{tan}\theta =-f\left(\theta \right)$
Equation (3) is the result of the odd function condition.

So,
$f\left(-a\right)=-f\left(a\right)=-2$
b) According to other property of tangent trigonometric function:
$\mathrm{tan}\theta =\mathrm{tan}\left(\theta +\pi \right)$
i.e. Period of  is $\pi$ which implies:
$\mathrm{tan}\theta =\mathrm{tan}\left(\theta +\pi \right)=\mathrm{tan}\left(\theta +2\pi \right)=\mathrm{tan}\left(\theta +3\pi \right)=\cdots =\mathrm{tan}\left(\theta +n\pi \right)$
Where n is an integer.
Hence from (1):
$f\left(\theta \right)=f\left(\theta +\pi \right)=f\left(\theta +2\pi \right)=f\left(\theta +3\pi \right)=\cdots =f\left(\theta +n\pi \right)$
Substitute $\theta =a:$
$f\left(a\right)=f\left(a+\pi \right)=f\left(a+2\pi \right)=f\left(a+3\pi \right)=\cdots =f\left(a+n\pi \right)$

Using (2):
$f\left(a\right)=f\left(a+\pi \right)=f\left(a+2\pi \right)=f\left(a+3\pi \right)=\cdots =f\left(a+n\pi \right)=2$
Then,
$f\left(a\right)+f\left(a+\pi \right)+f\left(a+2\pi \right)=3f\left(a\right)$
$3\cdot 2$
$=6$

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