Question

How would I solve this

Answer 1

sin∅ = √3/2

cos∅ = -1/2

Explanation:

Notice that 14π/3 is just under 3π, so the reference angle made with the negative x-axis is π/3. Since sin(π/3) = √3/2, then sin(14π/3) is also √3/2.

cos(14π/3) works somewhat similarly. Since cos(π/3) = 0.5, and cos(14π/3) would be located in Quadrant II where the negative axis is, then cos(14π/3) = -0.5

Answer 2

`\theta =\cfrac{14}{3}\pi \implies \theta =\cfrac{14\pi }{3}\implies \theta =\cfrac{6\pi +6\pi +2\pi }{3} \\\\\\ \theta =\cfrac{6\pi }{3}+\cfrac{6\pi }{3}+\cfrac{2\pi }{3}\implies \theta =2\pi +2\pi +\cfrac{2\pi }{3}`

so if we take a look at that, we can say that the angle θ does two revolutions and then it lands on 2π/3, so the terminal point of it is the same as 2π/3's, and if you check your Unit Circle, as you should have one

`\sin(\theta )=\cfrac{√(3)}{2}\hspace{5em}\cos(\theta )=-\cfrac{1}{2}`