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

  • Thread starter 01
  • Start date
  • Tags
    Cos Tan
In summary, to evaluate csc(theta) given cos(theta) = 1/2 and tan(theta) = negative, we can use the unit circle and CAST rule to determine that theta lies in the first or fourth quadrant, with possible values of pi/3 or 5pi/3. From there, we can use the definition of csc(theta) to find the value of -2sqrt[3]/3. Additionally, for the problem involving 3sin(3x+(pi/2)), the phase shift is pi/2, as determined by the general formula for phase shift in sinusoidal functions.
  • #1
01
10
0
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?
 
Physics news on Phys.org
  • #2
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
 
  • #3
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
 
Last edited:
  • #4
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.
 
  • #5
nevermind, i got it.
 

1. What are the primary trigonometric functions?

The primary trigonometric functions are cosine (cos), tangent (tan), and cosecant (csc). These functions are used to relate the angles of a right triangle to the lengths of its sides.

2. How do I solve a cos, tan, csc problem?

To solve a cos, tan, csc problem, you will need to use the given angle and side lengths to determine the trigonometric ratios for the triangle. Then, you can use these ratios to find the missing angle or side length.

3. What is the difference between cos, tan, and csc?

Cosine (cos) is the ratio of the adjacent side to the hypotenuse in a right triangle. Tangent (tan) is the ratio of the opposite side to the adjacent side in a right triangle. Cosecant (csc) is the reciprocal of the sine function, which is the ratio of the hypotenuse to the opposite side in a right triangle.

4. How do I remember the trigonometric ratios?

There are many mnemonic devices that can help you remember the trigonometric ratios. One popular method is the acronym "SOHCAHTOA", which stands for "Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent".

5. What are some real-life applications of cos, tan, and csc?

Trigonometric functions are used in many fields such as engineering, physics, and navigation. For example, cos and sin functions are used in calculating the trajectory of a projectile, while tan is used in determining the slope of a road or ramp. Cosecant is also used in designing bridges and buildings to ensure stability and structural integrity.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
262
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
194
  • Introductory Physics Homework Help
Replies
2
Views
630
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
907
  • Introductory Physics Homework Help
Replies
2
Views
975
  • Introductory Physics Homework Help
Replies
1
Views
890
Back
Top