(1 point) Given sec(πΌ)=23βΎβ/3 and πΌ is in quadrant I, find the exact values of the remaining five trigonometric functions.
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I suspect a typo, as before. I'll assume you meant that secπΌ = 2/β3
If you draw your triangle in standard position, then
r = 2
y = 1
x = β3
and then as usual,
sinπΌ = y/r
cosπΌ = x/r
tanπΌ = y/x
and so on.
To find the exact values of the remaining five trigonometric functions, we can use the given information that sec(πΌ) = 23βΎβ/3.
Since sec(πΌ) = 1/cos(πΌ), we can find the value of cos(πΌ) first.
sec(πΌ) = 23βΎβ/3
1/cos(πΌ) = 23βΎβ/3
cos(πΌ) = 3/(23βΎβ)
Using the Pythagorean identity sin^2(πΌ) + cos^2(πΌ) = 1, we can find the value of sin(πΌ).
sin^2(πΌ) + cos^2(πΌ) = 1
sin^2(πΌ) + (3/(23βΎβ))^2 = 1
sin^2(πΌ) + 9/23 = 1
sin^2(πΌ) = 1 - 9/23
sin^2(πΌ) = (23 - 9)/23
sin^2(πΌ) = 14/23
sin(πΌ) = β(14/23)
sin(πΌ) = β14/β23
Now, we can find the remaining trigonometric functions:
1. cos(πΌ) = 3/(23βΎβ)
2. tan(πΌ) = sin(πΌ) / cos(πΌ) = (β14/β23) / (3/(23βΎβ))
3. csc(πΌ) = 1/sin(πΌ) = 1 / (β14/β23)
4. sec(πΌ) = 1/cos(πΌ) = 1 / (3/(23βΎβ))
5. cot(πΌ) = 1/tan(πΌ) = 1 / [(β14/β23) / (3/(23βΎβ))]
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(πΌ) = 23β»β/3, we know that sec(πΌ) is the reciprocal of cos(πΌ). So, we can find cos(πΌ) by taking the reciprocal of sec(πΌ):
cos(πΌ) = 1/sec(πΌ) = 1/(23β»β/3).
Step 2: To simplify 1/(23β»β/3), we can rationalize the denominator. Multiply both the numerator and denominator by the conjugate of the denominator:
cos(πΌ) = (1*3)/(23β»β*3) = 3/(23β»β*3) = 3β3/(23β»β*3*β3) = 3β3/(23β»β*β9) = 3β3/(23β»β*β9) = 3β3/(23β»β*β9) = 3β3/(2β23).
So, cos(πΌ) = 3β3/(2β23).
Step 3: Next, we can find sin(πΌ) using the Pythagorean identity, sinΒ²(πΌ) + cosΒ²(πΌ) = 1. Since we have cos(πΌ), we can solve for sin(πΌ):
sinΒ²(πΌ) + (3β3/(2β23))^2 = 1
sinΒ²(πΌ) + (9*3)/(4*23) = 1
sinΒ²(πΌ) + 27/92 = 1
sinΒ²(πΌ) = 1 - 27/92
sinΒ²(πΌ) = (92 - 27)/92
sinΒ²(πΌ) = 65/92
sin(πΌ) = Β±β(65/92) = Β±β(13/23)
Since πΌ is in quadrant I, sin(πΌ) is positive. Therefore, sin(πΌ) = β(13/23).
Now, we have found cos(πΌ) = 3β3/(2β23) and sin(πΌ) = β(13/23).
Step 4: We can find the remaining trigonometric functions using the definitions and reciprocal relationships.
Tan(πΌ) = sin(πΌ)/cos(πΌ)
tan(πΌ) = (β(13/23)) / (3β3/(2β23))
tan(πΌ) = (2β23β(13/23))/(3β3)
tan(πΌ) = (2β13)/(3β3)
tan(πΌ) = 2β13/(3β3)
Csc(πΌ) = 1/sin(πΌ)
csc(πΌ) = 1/(β(13/23))
cosec(πΌ) = β(23/13) / (β(13/23) * β(23/13))
cosec(πΌ) = β23/β13
cosec(πΌ) = β23/β13 * (β13/β13)
cosec(πΌ) = β(23*13)/(β(13*13))
cosec(πΌ) = β299/β169
cosec(πΌ) = β299/13
Sec(πΌ) = 1/cos(πΌ)
sec(πΌ) = 1/(3β3/(2β23))
sec(πΌ) = (2β23)/(3β3) * (β3/β3)
sec(πΌ) = (2β23β3)/(3β3β3)
sec(πΌ) = (2β69)/(3*3)
sec(πΌ) = (2β69)/9
Cot(πΌ) = 1/tan(πΌ)
cot(πΌ) = 1/(2β13/(3β3))
cot(πΌ) = (3β3)/(2β13)
cot(πΌ) = (3β3)/(2β13) * (β13/β13)
cot(πΌ) = (3β3β13)/(2β13β13)
cot(πΌ) = (3β39)/(2*13)
cot(πΌ) = (3β39)/26
Hence, the exact values of the remaining five trigonometric functions are:
tan(πΌ) = 2β13/(3β3)
cosec(πΌ) = β299/13
sec(πΌ) = (2β69)/9
cot(πΌ) = (3β39)/26