therefore, cos(-270°) = cos(90) = 0 ans.
5. Find the exact value of sec 300 degrees
draw a Cartesian Coordinate similar to No. 4. you will notice that sec 300° = sec(-60°)
now draw a 30°-60°-90° triangle (this looks like a half of an equilateral triangle) with sides measuring 2, 1, and sqrt(3). take note that 2 is opposite the 90° angle, 1 is opposite the 30° angle and sqrt(3) is opposite the 60° angle. it is now apparent that
sec 300° = sec (-60) = 2/1
sec 300 = 2 ans.
6. Find the value of csc for angle A in standard position if the point at (5,-2) lies on its terminal side.
By the Distance Formula, the distance r from the origin (0,0) to any point (a,b) is
r = sqrt[(a - 0)^2 + (b - 0)^2]
r = sqrt(a^2 + b^2)
hence for point (5,-2), we have
r = sqrt[5^2 + (-2)^2]
r = sqrt(29) = 5.39
now draw a right triangle with a = 5, b = -2, and r = 5.39 (this is the hypotenuse0
cscA = r/a
cscA = 5.39/5
cscA = 1.078 ans.
7. Suppose (A) is an angle in standard position whose terminal side lies in quadrant . If sin(A)=12/13,find the value of sec(A)
note: your question does not indicate in which quadrant the terminal side of angle A lies. a positive value of sinA can only be in quadrant I or quadrant II.
solve first for the missing side b, since sinA = a/c = 12/13
using Pythagorean Theorem,
b^2 = c^2 - a^2
b^2 = 13^2 - 12^2
b^2 = 169 - 144
b^2 = 25
b = 5
for quadrant I,
secA = 13/5 ans.
for quadrant II,
b = -5
secA = 13/(-5) ans