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Section 1.3 : Trig Functions

9. Determine the exact value of \(\displaystyle \tan \left( {\frac{{15\pi }}{4}} \right)\) without using a calculator.

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Hint : Even though a unit circle only tells us information about sine and cosine it is still useful for tangents so sketch a unit circle and relate the angle to one of the standard angles in the first quadrant.
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First we can notice that \(4\pi - \frac{\pi }{4} = \frac{{15\pi }}{4}\) and also note that \(4\pi \) is two complete revolutions so the terminal line for \(\frac{{15\pi }}{4}\) and \( - \frac{\pi }{4}\) represent the same angle. Also note that \( - \frac{\pi }{4}\) will form an angle of \(\frac{\pi }{4}\) with the positive \(x\)-axis in the fourth quadrant and we’ll have the following unit circle for this problem.

TrigFcns_Prob9
Hint : Given the obvious symmetry in the unit circle relate the coordinates of the line representing \(\frac{{15\pi }}{4}\) to the coordinates of the line representing \(\frac{\pi }{4}\) and the definition of tangent in terms of sine and cosine to answer the question.
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The coordinates of the line representing \(\frac{{15\pi }}{4}\) will be the same as the coordinates of the line representing \(\frac{\pi }{4}\) except that the \(y\) coordinate will now be negative. So, our new coordinates will then be \(\left( {\frac{{\sqrt 2 }}{2}, - \frac{{\sqrt 2 }}{2}} \right)\) and so the answer is,

\[\tan \left( \frac{15\pi }{4} \right)=\frac{\sin \left( \frac{15\pi }{4} \right)}{\cos \left( \frac{15\pi }{4} \right)}=\frac{-{}^{\sqrt{2}}/{}_{2}}{{}^{\sqrt{2}}/{}_{2}}=-1\]