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Alla [95]
4 years ago
13

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

Mathematics
1 answer:
Daniel [21]4 years ago
8 0
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
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