Practice: Trigonometry

(a) $\sin \frac{7\pi }{6}$ and $\cos \frac{7\pi }{6}$. $\frac{7\pi }{6}=210°$, which lies in the third quadrant, past $\pi$ but before $3\pi /2$. There both coordinates are negative. The reference angle is $\frac{7\pi }{6}-\pi =\frac{\pi }{6}$, whose values come from the 30-60-90 triangle: $\sin \frac{\pi }{6}=\frac{1}{2}$ and $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$. Applying the third-quadrant signs:

Check: $\sin (7\pi /6)=-0.5000000000$ ✓. The quadrant supplies the signs and the reference angle supplies the magnitudes. No separate sign rule is needed, because the point's position determines both. (b) $\cos t$ and $\tan t$ given $\sin t=-\frac{5}{13}$, $t$ in the fourth quadrant. From the Pythagorean identity:

The identity gives only the magnitude. In the fourth quadrant the first coordinate is positive, so

Then

Check: ${\left(\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2=\frac{25+144}{169}=1$ ✓. The negative tangent is consistent: in the fourth quadrant sine is negative and cosine positive, so their quotient is negative. (c) $\sin \frac{\pi }{12}$ by angle addition. $\frac{\pi }{12}=15°=45°-30°$, a difference of two special angles. Using $\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$:

Check: $\sin (\pi /12)=0.258819045103$ and $(\sqrt{6}-\sqrt{2})/4=0.258819045103$ ✓. The table has five entries; angle addition reaches every sum or difference of them, so $15°$, $75°$, $105°$ and so on all acquire exact values. (d) $\sin x=-\frac{1}{2}$ on $[0,2\pi )$, and in general. The magnitude $\frac{1}{2}$ gives reference angle $\frac{\pi }{6}$. Sine is the second coordinate, negative in the third and fourth quadrants: | Quadrant | Solution |
|---|---|
| third | $\pi +\frac{\pi }{6}=\frac{7\pi }{6}$ |
| fourth | $2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$ | Check: $\sin (7\pi /6)=-0.5000000000$ and $\sin (11\pi /6)=-0.5000000000$ ✓. General solution. Sine has period $2\pi$, so add whole revolutions to each:

The solution set is infinite; the interval $[0,2\pi )$ is what makes the answer two values rather than infinitely many. (e) Undefined points, the definition, and the angle unit. Where $\tan \theta$ is undefined. $\tan \theta =\frac{\sin \theta }{\cos \theta }$, so it is undefined exactly where $\cos \theta =0$, at

These are the points where the unit circle crosses the vertical axis, so the first coordinate vanishes and the quotient divides by zero. The function has no value there, rather than an infinite one: the computed $\cos (\pi /2)$ is $6.1\times {10}^{-17}$, and nearby tangents climb through $\mathrm{10,381}$ at $1.5707$ and $\mathrm{3,060,023}$ at $1.570796$ without ever reaching a value at $\pi /2$ itself. Why the circle definition supersedes the triangle one. A right triangle's angles are acute and its side lengths positive, so the ratio definition can only produce values for $\theta \in (0,\pi /2)$ and only positive ones. It cannot express $\cos \frac{2\pi }{3}=-\frac{1}{2}$, because no right triangle has a $120°$ angle, and it cannot express a negative or a reflex angle at all. The circle definition reads coordinates, which carry signs, so every real $\theta$, negative, obtuse, beyond a full revolution, names a point and therefore has a sine and a cosine. The two definitions agree wherever both apply: for acute $\theta$, the point $(\cos \theta ,\sin \theta )$ is the corner of a right triangle with hypotenuse 1, so the coordinates are the ratios. The circle extends the triangle rather than replacing it. What breaks in degrees. The $\theta$ in the definition is an arc length on a circle of radius 1. A pure number. Degrees are an arbitrary subdivision, and using them inserts a constant into every limit and derivative:

Since that limit is what establishes $\frac{d}{dx}\sin x=\cos x$, the derivative formula is false in degrees. It becomes $\frac{\pi }{180}\cos x$. Every subsequent formula would carry the same factor, which is why analysis fixes radians and treats degrees as a conversion for input and output only.