(1 point) Given sec(𝛼)=23‾√/3 and 𝛼 is in quadrant I, find the exact values of the remaining five trigonometric functions.

Question

(1 point) Given sec(𝛼)=23‾√/3 and 𝛼 is in quadrant I, find the exact values of the remaining five trigonometric functions.

Note: You are not allowed to use decimals in your answer.

Answer 1

I suspect a typo, as before. I'll assume you meant that sec𝛼 = 2/√3

If you draw your triangle in standard position, then
r = 2
y = 1
x = √3
and then as usual,
sin𝛼 = y/r
cos𝛼 = x/r
tan𝛼 = y/x
and so on.

Answer 2

To find the exact values of the remaining five trigonometric functions, we can use the given information that sec(𝛼) = 23‾√/3.

Since sec(𝛼) = 1/cos(𝛼), we can find the value of cos(𝛼) first.

sec(𝛼) = 23‾√/3
1/cos(𝛼) = 23‾√/3
cos(𝛼) = 3/(23‾√)

Using the Pythagorean identity sin^2(𝛼) + cos^2(𝛼) = 1, we can find the value of sin(𝛼).

sin^2(𝛼) + cos^2(𝛼) = 1
sin^2(𝛼) + (3/(23‾√))^2 = 1
sin^2(𝛼) + 9/23 = 1
sin^2(𝛼) = 1 - 9/23
sin^2(𝛼) = (23 - 9)/23
sin^2(𝛼) = 14/23
sin(𝛼) = √(14/23)
sin(𝛼) = √14/√23

Now, we can find the remaining trigonometric functions:

1. cos(𝛼) = 3/(23‾√)
2. tan(𝛼) = sin(𝛼) / cos(𝛼) = (√14/√23) / (3/(23‾√))
3. csc(𝛼) = 1/sin(𝛼) = 1 / (√14/√23)
4. sec(𝛼) = 1/cos(𝛼) = 1 / (3/(23‾√))
5. cot(𝛼) = 1/tan(𝛼) = 1 / [(√14/√23) / (3/(23‾√))]

Answer 3

To find the exact values of the remaining five trigonometric functions, we can use the reciprocal relationships and special right triangles. Here's the step-by-step process:

Step 1: Given sec(𝛼) = 23⁻√/3, we know that sec(𝛼) is the reciprocal of cos(𝛼). So, we can find cos(𝛼) by taking the reciprocal of sec(𝛼):
cos(𝛼) = 1/sec(𝛼) = 1/(23⁻√/3).

Step 2: To simplify 1/(23⁻√/3), we can rationalize the denominator. Multiply both the numerator and denominator by the conjugate of the denominator:
cos(𝛼) = (1*3)/(23⁻√*3) = 3/(23⁻√*3) = 3√3/(23⁻√*3*√3) = 3√3/(23⁻√*√9) = 3√3/(23⁻√*√9) = 3√3/(23⁻√*√9) = 3√3/(2√23).

So, cos(𝛼) = 3√3/(2√23).

Step 3: Next, we can find sin(𝛼) using the Pythagorean identity, sin²(𝛼) + cos²(𝛼) = 1. Since we have cos(𝛼), we can solve for sin(𝛼):
sin²(𝛼) + (3√3/(2√23))^2 = 1
sin²(𝛼) + (9*3)/(4*23) = 1
sin²(𝛼) + 27/92 = 1
sin²(𝛼) = 1 - 27/92
sin²(𝛼) = (92 - 27)/92
sin²(𝛼) = 65/92
sin(𝛼) = ±√(65/92) = ±√(13/23)

Since 𝛼 is in quadrant I, sin(𝛼) is positive. Therefore, sin(𝛼) = √(13/23).

Now, we have found cos(𝛼) = 3√3/(2√23) and sin(𝛼) = √(13/23).

Step 4: We can find the remaining trigonometric functions using the definitions and reciprocal relationships.

Tan(𝛼) = sin(𝛼)/cos(𝛼)
tan(𝛼) = (√(13/23)) / (3√3/(2√23))
tan(𝛼) = (2√23√(13/23))/(3√3)
tan(𝛼) = (2√13)/(3√3)
tan(𝛼) = 2√13/(3√3)

Csc(𝛼) = 1/sin(𝛼)
csc(𝛼) = 1/(√(13/23))
cosec(𝛼) = √(23/13) / (√(13/23) * √(23/13))
cosec(𝛼) = √23/√13
cosec(𝛼) = √23/√13 * (√13/√13)
cosec(𝛼) = √(23*13)/(√(13*13))
cosec(𝛼) = √299/√169
cosec(𝛼) = √299/13

Sec(𝛼) = 1/cos(𝛼)
sec(𝛼) = 1/(3√3/(2√23))
sec(𝛼) = (2√23)/(3√3) * (√3/√3)
sec(𝛼) = (2√23√3)/(3√3√3)
sec(𝛼) = (2√69)/(3*3)
sec(𝛼) = (2√69)/9

Cot(𝛼) = 1/tan(𝛼)
cot(𝛼) = 1/(2√13/(3√3))
cot(𝛼) = (3√3)/(2√13)
cot(𝛼) = (3√3)/(2√13) * (√13/√13)
cot(𝛼) = (3√3√13)/(2√13√13)
cot(𝛼) = (3√39)/(2*13)
cot(𝛼) = (3√39)/26

Hence, the exact values of the remaining five trigonometric functions are:
tan(𝛼) = 2√13/(3√3)
cosec(𝛼) = √299/13
sec(𝛼) = (2√69)/9
cot(𝛼) = (3√39)/26