how would I solve this

How Would I Solve This

Answers

Answer 1

[tex]\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}[/tex]

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

[tex]\sin(\theta )=\cfrac{\sqrt{3}}{2}\hspace{5em}\cos(\theta )=-\cfrac{1}{2}[/tex]

Answer 2

Answer:

sin∅ = √3/2

cos∅ = -1/2

Step-by-step 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


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