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

How do i solve this trigonometry question? (Section 7.3)-example-1
User RGS
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1 Answer

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Solve:
tan(-(\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(-(\pi)/(12)) = -tan((\pi)/(12))

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


(\pi)/(3) - (\pi)/(4) = (4\pi - 3\pi)/(12) = (\pi)/(12)

Thus, we can say:

-tan((\pi)/(12)) = -tan((\pi)/(3) - (\pi)/(4))

Using difference formula to simplify the expression, we know that the difference formula states:

tan(A - B) = (tanA - tanB)/(1 + tanA \cdot tanB)

-tan((\pi)/(3) - (\pi)/(4)) = -(tan((\pi)/(3) - tan((\pi)/(4)))/(1 + tan((\pi)/(3)) \cdot tan((\pi)/(4)))

The exact value for
tan((\pi)/(3)) = √(3)
The exact value for
tan((\pi)/(4)) = 1

Thus:

-(tan((\pi)/(3) - tan((\pi)/(4)))/(1 + tan((\pi)/(3)) \cdot tan((\pi)/(4))) = -(√(3) - 1)/(1 + √(3))

= (1 - √(3))/(1 + √(3))

Rationalise the denominator:

(1 - √(3))/(1 + √(3)) \cdot (1 - √(3))/(1 - √(3))

= ((1 - √(3))^(2))/(1 - 3)

Expand the perfect square:

((1 - √(3))^(2))/(1 - 3) = (1 - 2√(3) + 3)/(-2)

= (4 - 2√(3))/(-2)

= -2 + √(3)


\therefore &nbsp;<span>tan(-(\pi)/(12)) = √(3) - 2</span>
User Dmitrii Bychkov
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