Cos, Tan, and Csc Problem: Evaluating Trigonometric Functions for a Given Theta

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To evaluate csc(theta) given cos(theta) = 1/2 and tan(theta) is negative, the discussion highlights that theta must be in the fourth quadrant. The angles corresponding to cos(theta) = 1/2 are pi/3 and 5pi/3, but since tan(theta) is negative, theta is determined to be 5pi/3. Consequently, csc(theta) is calculated as -2√3/3. The thread also briefly touches on a separate question about the phase shift of a sine function, which the user resolves independently.
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doing a problem but can't 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?
 
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01 said:
doing a problem but can't 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
 
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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.
 
nevermind, i got it.
 
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