23.8k views
0 votes
Find the exact value of the trigonometric function at the given real number.

(a) cos(19/6)
(b) cos(−7/6
(c) cos(−11/6).

User DengApro
by
7.7k points

1 Answer

4 votes

Final Answer:

(a) cos(19/6) = 1/2

(b) cos(-7/6) = √3/2

(c) cos(-11/6) = -√3/2

Step-by-step explanation:

Trigonometric functions are periodic, and their values repeat after a certain interval. To find the exact values of cos(19/6), cos(-7/6), and cos(-11/6), we can use the unit circle and the periodic nature of cosine.

Starting with (a) cos(19/6), we note that 19/6 is equivalent to 3π + π/6. On the unit circle, at 3π + π/6, the cosine value is 1/2. Therefore, cos(19/6) = 1/2.

Moving on to (b) cos(-7/6), -7/6 is equivalent to -π + π/6. On the unit circle, at -π + π/6, the cosine value is √3/2. Hence, cos(-7/6) = √3/2.

Finally, for (c) cos(-11/6), -11/6 is equivalent to -2π - π/6. On the unit circle, at -2π - π/6, the cosine value is also √3/2, but negative since it's in the third quadrant. Therefore, cos(-11/6) = -√3/2.

In summary, using the unit circle and recognizing the periodicity of cosine, we determined that cos(19/6) = 1/2, cos(-7/6) = √3/2, and cos(-11/6) = -√3/2.

User Vintnes
by
7.9k points