Find the exact value of the cosine and the cosecant of thetha, if thetha terminates in quadrant II and sin theta = 4/5
cos^2 + sin^2 =1
cos^2 = 1-16/25 =9/25
sin theta terminates in quadrant II, so sin>0 but cos <0
therefore cos =-3/5
To find the exact value of the cosine and cosecant of theta, we need to use the given information that theta terminates in quadrant II and sin(theta) = 4/5.
In quadrant II, both the x-coordinate and y-coordinate of a point on the unit circle are negative. Since sin(theta) = 4/5 is positive, we can conclude that the y-coordinate (opposite side) is 4, and the x-coordinate (adjacent side) is negative.
To find the cosine of theta, we can use the Pythagorean identity: cos^2(theta) + sin^2(theta) = 1. Since sin(theta) = 4/5, we can substitute this value into the equation:
cos^2(theta) + (4/5)^2 = 1
cos^2(theta) + 16/25 = 1
cos^2(theta) = 1 - 16/25
cos^2(theta) = 25/25 - 16/25
cos^2(theta) = 9/25
cos(theta) = sqrt(9/25)
cos(theta) = 3/5
Therefore, the exact value of the cosine of theta in quadrant II is 3/5.
To find the cosecant of theta, we can use the reciprocal of sin(theta):
cosec(theta) = 1/sin(theta)
cosec(theta) = 1/(4/5)
cosec(theta) = 5/4
Therefore, the exact value of the cosecant of theta in quadrant II is 5/4.