the following are equivalent sine ratios for one full standard angle rotation around an unit circle…

the following are equivalent sine ratios for one full standard angle rotation around an unit circle: \n$\\sin ( \\frac { \\pi } { 12 } ) = \\sin ( \\frac { 11 \\pi } { 12 } ) = \\sin ( \\frac { 13 \\pi } { 12 } ) = \\sin ( \\frac { 23 \\pi } { 12 } )$\ntrue\nfalse

the following are equivalent sine ratios for one full standard angle rotation around an unit circle: \n$\\sin ( \\frac { \\pi } { 12 } ) = \\sin ( \\frac { 11 \\pi } { 12 } ) = \\sin ( \\frac { 13 \\pi } { 12 } ) = \\sin ( \\frac { 23 \\pi } { 12 } )$\ntrue\nfalse

Answer

Explanation:

Step1: Use the sine function property

The sine function has the property (\sin(\pi - \alpha)=\sin\alpha) and (\sin(2\pi-\alpha)=-\sin\alpha), (\sin(\alpha + 2k\pi)=\sin\alpha,k\in\mathbb{Z}). For (\sin\left(\frac{11\pi}{12}\right)), we have (\sin\left(\frac{11\pi}{12}\right)=\sin\left(\pi-\frac{\pi}{12}\right)). According to the formula (\sin(\pi - \alpha)=\sin\alpha), so (\sin\left(\pi-\frac{\pi}{12}\right)=\sin\frac{\pi}{12}). For (\sin\left(\frac{13\pi}{12}\right)), we have (\sin\left(\frac{13\pi}{12}\right)=\sin\left(\pi+\frac{\pi}{12}\right)). According to the formula (\sin(\pi+\alpha)=-\sin\alpha), so (\sin\left(\pi + \frac{\pi}{12}\right)=-\sin\frac{\pi}{12}). For (\sin\left(\frac{23\pi}{12}\right)), we have (\sin\left(\frac{23\pi}{12}\right)=\sin\left(2\pi-\frac{\pi}{12}\right)). According to the formula (\sin(2\pi-\alpha)=-\sin\alpha), so (\sin\left(2\pi-\frac{\pi}{12}\right)=-\sin\frac{\pi}{12}).

Answer:

False