Question

How do i solve this trigonometry question? (Section 7.3)

Answer 1

Solve: $tan(-\frac{\pi}{12})$

Now, a tangent function is an odd function, and you can find the proof on the internet, so we can say that:
$f(-x) = -f(x)$
$tan(-x) = -tan(x)$
$tan(-\frac{\pi}{12}) = -tan(\frac{\pi}{12})$

Now, we just need to find what $tan(\frac{\pi}{12})$ is.
Since $\frac{\pi}{12}$ is not an exact angle, we need to manipulate it in some way that it will be yield exact values.

$\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$

Thus, we can say:
$-tan(\frac{\pi}{12}) = -tan(\frac{\pi}{3} - \frac{\pi}{4})$

Using difference formula to simplify the expression, we know that the difference formula states:
$tan(A - B) = \frac{tanA - tanB}{1 + tanA \cdot tanB}$
$-tan(\frac{\pi}{3} - \frac{\pi}{4}) = -\frac{tan(\frac{\pi}{3} - tan(\frac{\pi}{4})}{1 + tan(\frac{\pi}{3}) \cdot tan(\frac{\pi}{4})}$

The exact value for $tan(\frac{\pi}{3}) = \sqrt{3}$
The exact value for $tan(\frac{\pi}{4}) = 1$

Thus:
$-\frac{tan(\frac{\pi}{3} - tan(\frac{\pi}{4})}{1 + tan(\frac{\pi}{3}) \cdot tan(\frac{\pi}{4})} = -\frac{\sqrt{3} - 1}{1 + \sqrt{3}}$
$= \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Rationalise the denominator:
$\frac{1 - \sqrt{3}}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$
$= \frac{(1 - \sqrt{3})^{2}}{1 - 3}$

Expand the perfect square:
$\frac{(1 - \sqrt{3})^{2}}{1 - 3} = \frac{1 - 2\sqrt{3} + 3}{-2}$
$= \frac{4 - 2\sqrt{3}}{-2}$
$= -2 + \sqrt{3}$

$\therefore tan(-\frac{\pi}{12}) = \sqrt{3} - 2$