Answer:
1) False
2) True
3) False
Step-by-step explanation:
From inspection of the attached unit circle, we can see that:
[tex]\boxed{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}[/tex]
Therefore, the first statement is false.
[tex]\hrulefill[/tex]
The tangent function is periodic with a period of π, meaning that for any integer n:
[tex]\tan(a)=\tan(a+\pi n)[/tex]
[tex]\tan(n\pi)=\tan \pi[/tex]
Therefore:
[tex]\tan(27\pi)=\tan \pi[/tex]
From the unit circle we can see that:
[tex]\tan \pi=\dfrac{\sin \pi}{\cos \pi}=\dfrac{0}{-1}=0[/tex]
Therefore,
[tex]\boxed{\tan(27\pi)=0}[/tex]
Therefore, the second statement is true.
[tex]\hrulefill[/tex]
[tex]\textsf{Using the trigonometric identity\; $\sec x=\dfrac{1}{\cos x}$}:[/tex]
[tex]\sec \left(-\dfrac{11\pi}{3}\right)=\dfrac{1}{\cos \left(-\dfrac{11\pi}{3}\right)}[/tex]
The cosine function is periodic with a period of 2π, meaning that for any integer n:
[tex]\cos (a)=\cos (a + 2\pi n)[/tex]
Therefore:
[tex]\cos \left(-\dfrac{11\pi}{3}\right)=\cos \left(-\dfrac{11\pi}{3}+4\pi\right)=\cos \left(\dfrac{\pi}{3}\right)[/tex]
From the unit circle we can see that:
[tex]\cos \left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}[/tex]
Therefore:
[tex]\sec \left(-\dfrac{11\pi}{3}\right)=\dfrac{1}{\cos \left(-\dfrac{11\pi}{3}\right)}=\dfrac{1}{\cos \left(\dfrac{\pi}{3}\right)}=\dfrac{1}{\frac{1}{2}}=2[/tex]
Therefore, the third statement is false.
