Question

4 multiple choice 1 point which of the following is tan(4π/6)? -√3 -√3/2 -√3/3 -1/2 previous

Explanation:

Step1: Use the tangent subtraction formula

We know that \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\). Also, \(\frac{4\pi}{6}=\pi-\frac{\pi}{3}\), and \(\tan(\pi - \alpha)=-\tan\alpha\). So \(\tan(\frac{4\pi}{6})=\tan(\pi-\frac{\pi}{3})\).
Since \(\tan(\pi - \alpha)=-\tan\alpha\), then \(\tan(\frac{4\pi}{6})=-\tan(\frac{\pi}{3})\). But another way is to use the unit - circle or reference - angle concept. \(\frac{4\pi}{6}=\frac{2\pi}{3}\) (simplified). The reference angle for \(\theta=\frac{4\pi}{6}\) (or \(\frac{2\pi}{3}\)) is \(\pi-\frac{4\pi}{6}=\frac{\pi}{3}\). And \(\tan\theta\) in the second quadrant (\(\frac{\pi}{2}<\theta<\pi\)) is negative. Also, \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), \(\sin(\frac{4\pi}{6})=\frac{\sqrt{3}}{2}\), \(\cos(\frac{4\pi}{6})=-\frac{1}{2}\), so \(\tan(\frac{4\pi}{6})=\frac{\sin(\frac{4\pi}{6})}{\cos(\frac{4\pi}{6})}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}\). Another approach: \(\tan(x)=\frac{\sin(x)}{\cos(x)}\), \(\sin(\frac{4\pi}{6})=\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\), \(\cos(\frac{4\pi}{6})=\cos(\frac{2\pi}{3})=-\frac{1}{2}\), \(\tan(\frac{4\pi}{6})=\frac{\sin(\frac{4\pi}{6})}{\cos(\frac{4\pi}{6})}\).

Step2: Calculate the value

\(\tan(\frac{4\pi}{6})=\frac{\sin(\frac{4\pi}{6})}{\cos(\frac{4\pi}{6})}\). We know that \(\sin(\frac{4\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\cos(\frac{4\pi}{6})=-\frac{1}{2}\). Then \(\tan(\frac{4\pi}{6})=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}\).

Another way: \(\tan(x)\) has a period of \(\pi\), \(\tan(\frac{4\pi}{6})=\tan(\frac{2\pi}{3})\). Using the formula \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\), \(\frac{2\pi}{3}=\frac{\pi}{2}+\frac{\pi}{6}\), but it's more complex. Using the unit - circle definition: for an angle \(\theta\) in standard position, if the terminal side of \(\theta\) intersects the unit circle at the point \((x,y)\), then \(\tan\theta=\frac{y}{x}\). For \(\theta = \frac{4\pi}{6}\), the point on the unit circle is \((-\frac{1}{2},\frac{\sqrt{3}}{2})\), so \(\tan(\frac{4\pi}{6})=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}\).

Answer:

\(-\sqrt{3}\)