View Full Version : cos, tan, csc problem
doing a problem but cant figure out the answer:
Given the following, evaluate csc(theta):
cos(theta) = 1/2
tan(theta) = negative
a> 2
b> -2(square root of)3/3
c> 2(square root of)3/3
d> -2
any help?
stunner5000pt
Dec10-04, 12:48 PM
doing a problem but cant figure out the answer:
Given the following, evaluate csc(theta):
cos(theta) = 1/2
tan(theta) = negative
a> 2
b> -2(square root of)3/3
c> 2(square root of)3/3
d> -2
any help?
i do not understand the part with the a b c and d in it but i can tel lyou that if you look at your unit circle (CAST) rule then you can fiure out the point where the cos is positive and the tan is negative
if cos theta = 0.5 and tan theta < 0 then theta can onlky lie in the first or fourth quadrant i.e. between 0 and 90 and 270 and 360
or if you care in radians 0 to pi/2 and 3pi/2 and 2pi
if the cos(theta) = 1/2, then (theta) = pi/3, or 5pi/3, on the interval [0, 2pi],
since it is given that tan(theta) = sin(theta)/cos(theta) < 0,
sin(theta)/(1/2) = 2sin(theta) < 0 or sin(theta) < 0,
therefore (theta) has to be (5pi/3)
now, csc(theta) = 1/sin(theta) = 1/(-3/(2sqrt[3]) = -2sqrt[3]/3
thx for the input, now here's another:
what is the phase shift for 3sin (3x+(pi/2))
a> -pi/6
b> 3
c> pi/2
d> -3pi/2
*the a,b,c,& d are multiple choices on the assignment I'm doing.
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