Question

use the information contained on the unit circle above to identify which of the following expressions are equivalent to one.
□ ( sin left( \frac { 5 pi } { 4 }
ight) + cos left( \frac { 5 pi } { 4 }
ight) )
□ ( sin left( \frac { 2 pi } { 3 }
ight) div cos left( \frac { pi } { 6 }
ight) )
□ ( - 2 cdot sin left( \frac { 5 pi } { 3 }
ight) )
□ ( \tan left( \frac { pi } { 6 }
ight) cdot \tan left( \frac { 5 pi } { 3 }
ight) )
□ ( sin left( \frac { pi } { 2 }
ight) cdot \tan left( \frac { 5 pi } { 4 }
ight) )

Explanation:

Step1: Recall unit - circle values

On the unit circle, for an angle \(\theta\), \(\sin\theta = y\) - coordinate and \(\cos\theta=x\) - coordinate of the point on the unit circle corresponding to \(\theta\), and \(\tan\theta=\frac{\sin\theta}{\cos\theta}\).

• For \(\theta = \frac{5\pi}{4}\), \(\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\), \(\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\). Then \(\sin(\frac{5\pi}{4})+\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}

eq1\).

• For \(\theta=\frac{2\pi}{3}\), \(\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\); for \(\theta = \frac{\pi}{6}\), \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\). Then \(\sin(\frac{2\pi}{3})\div\cos(\frac{\pi}{6})=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1\).
• For \(\theta=\frac{5\pi}{3}\), \(\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}\). Then \(- 2\cdot\sin(\frac{5\pi}{3})=-2\times(-\frac{\sqrt{3}}{2})=\sqrt{3}

eq1\).

• For \(\theta=\frac{\pi}{6}\), \(\tan(\frac{\pi}{6})=\frac{\sqrt{3}}{3}\); for \(\theta=\frac{5\pi}{3}\), \(\tan(\frac{5\pi}{3})=-\sqrt{3}\). Then \(\tan(\frac{\pi}{6})\cdot\tan(\frac{5\pi}{3})=\frac{\sqrt{3}}{3}\times(-\sqrt{3})=- 1

eq1\).

• For \(\theta=\frac{\pi}{2}\), \(\sin(\frac{\pi}{2}) = 1\); for \(\theta=\frac{5\pi}{4}\), \(\tan(\frac{5\pi}{4})=\frac{\sin(\frac{5\pi}{4})}{\cos(\frac{5\pi}{4})}=1\). Then \(\sin(\frac{\pi}{2})\cdot\tan(\frac{5\pi}{4})=1\times1 = 1\).

Answer:

\(\sin(\frac{2\pi}{3})\div\cos(\frac{\pi}{6})\), \(\sin(\frac{\pi}{2})\cdot\tan(\frac{5\pi}{4})\)