Solve: [tex]tan(-\frac{\pi}{12})[/tex]
Now, a tangent function is an odd function, and you can find the proof on the internet, so we can say that:
[tex]f(-x) = -f(x)[/tex]
[tex]tan(-x) = -tan(x)[/tex]
[tex]tan(-\frac{\pi}{12}) = -tan(\frac{\pi}{12})[/tex]
Now, we just need to find what [tex]tan(\frac{\pi}{12})[/tex] is.
Since [tex]\frac{\pi}{12}[/tex] is not an exact angle, we need to manipulate it in some way that it will be yield exact values.
[tex]\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}[/tex]
Thus, we can say:
[tex]-tan(\frac{\pi}{12}) = -tan(\frac{\pi}{3} - \frac{\pi}{4})[/tex]
Using difference formula to simplify the expression, we know that the difference formula states:
[tex]tan(A - B) = \frac{tanA - tanB}{1 + tanA \cdot tanB}[/tex]
[tex]-tan(\frac{\pi}{3} - \frac{\pi}{4}) = -\frac{tan(\frac{\pi}{3} - tan(\frac{\pi}{4})}{1 + tan(\frac{\pi}{3}) \cdot tan(\frac{\pi}{4})}[/tex]
The exact value for [tex]tan(\frac{\pi}{3}) = \sqrt{3}[/tex]
The exact value for [tex]tan(\frac{\pi}{4}) = 1[/tex]
Thus:
[tex]-\frac{tan(\frac{\pi}{3} - tan(\frac{\pi}{4})}{1 + tan(\frac{\pi}{3}) \cdot tan(\frac{\pi}{4})} = -\frac{\sqrt{3} - 1}{1 + \sqrt{3}}[/tex]
[tex]= \frac{1 - \sqrt{3}}{1 + \sqrt{3}}[/tex]
Rationalise the denominator:
[tex]\frac{1 - \sqrt{3}}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}}[/tex]
[tex]= \frac{(1 - \sqrt{3})^{2}}{1 - 3}[/tex]
Expand the perfect square:
[tex]\frac{(1 - \sqrt{3})^{2}}{1 - 3} = \frac{1 - 2\sqrt{3} + 3}{-2}[/tex]
[tex]= \frac{4 - 2\sqrt{3}}{-2}[/tex]
[tex]= -2 + \sqrt{3}[/tex]
[tex]\therefore tan(-\frac{\pi}{12}) = \sqrt{3} - 2[/tex]
