Can you help me with this trig homework question?
2016-04-27 7:35 am
use the function value given to determine the value of the other five trig functions of the acute angle theta. Answer in exact form. Tan theta <0 and Cos theta=5/8
回答 (2)
2016-04-27 7:49 am
Since cosθ > 0 and tanθ < 0
then θ is in the 4th quadrant, and sinθ < 0
As sinθ < 0
sin²θ + cos²θ = 1
sin²θ = 1 - cos²θ
sinθ = -√(1 - cos²θ)
sinθ = -√[1 - (5/8)²]
sinθ = -√(39)/64
sinθ = -(√39)/8
tanθ = sinθ / cosθ
tanθ = [-(√39)/8] / (5/8)
tanθ = -(√39)/5
secθ = 1 / cosθ
secθ = 1 / (5/8)
secθ = 8/5
cscθ = 1 / sinθ
cscθ = 1 / [-(√39)/8]
csc = -8/(√39)
csc = -8/(√39) × [(√39)/(√39)]
csc = -8(√39)/39
cotθ = 1 / tanθ
cotθ = 1 / [-(√39)/5]
cotθ = -5/(√39)
cotθ = -5/(√39) × [(√39)/(√39)]
cotθ = -5(√39)/39
then θ is in the 4th quadrant, and sinθ < 0
As sinθ < 0
sin²θ + cos²θ = 1
sin²θ = 1 - cos²θ
sinθ = -√(1 - cos²θ)
sinθ = -√[1 - (5/8)²]
sinθ = -√(39)/64
sinθ = -(√39)/8
tanθ = sinθ / cosθ
tanθ = [-(√39)/8] / (5/8)
tanθ = -(√39)/5
secθ = 1 / cosθ
secθ = 1 / (5/8)
secθ = 8/5
cscθ = 1 / sinθ
cscθ = 1 / [-(√39)/8]
csc = -8/(√39)
csc = -8/(√39) × [(√39)/(√39)]
csc = -8(√39)/39
cotθ = 1 / tanθ
cotθ = 1 / [-(√39)/5]
cotθ = -5/(√39)
cotθ = -5/(√39) × [(√39)/(√39)]
cotθ = -5(√39)/39
2016-04-27 8:12 am
Acute angle, tan θ < 0 ??? Take it as negative acute (?)
cos θ = 5/8
sin^2 θ = 1 - cos^2 θ = 39/64
tan θ < 0, so sin θ < 0, too.
So sin θ = -√39 / 8
tan θ = sin θ / cos θ = -√39 / 5
sec θ = 1 / cos θ = 8/5
csc θ = 1 / sin θ = -8 / √39, = -8 √39 / 39
cot θ = 1 / tam θ = -5 / √39, = -5 √39 / 39
cos θ = 5/8
sin^2 θ = 1 - cos^2 θ = 39/64
tan θ < 0, so sin θ < 0, too.
So sin θ = -√39 / 8
tan θ = sin θ / cos θ = -√39 / 5
sec θ = 1 / cos θ = 8/5
csc θ = 1 / sin θ = -8 / √39, = -8 √39 / 39
cot θ = 1 / tam θ = -5 / √39, = -5 √39 / 39
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