PLEASE HELP!!!!

Mathematics

2 answers:

kherson [118]2 years ago

5 0

Answers:

sin(θ) = -15/17
cos(θ) = 8/17
tan(θ) = -15/8
csc(θ) = -17/15
sec(θ) = 17/8
cot(θ) = -8/15

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Explanation:

Refer to the diagram below. The point (8,-15) is in quadrant 4, which is the southeast quadrant. A right triangle forms where the hypotenuse goes from the origin to the terminal point in question. After applying the pythagorean theorem, the hypotenuse is 17 units long.

Opposite the reference angle θ, greek letter theta, the side is -15. Usually negative lengths are not allowed and don't make any sense; however, I'm making this negative to ensure that sin(θ) ends up negative. This will help with other trig ratios as well.

So,

sin(angle) = opposite/hypotenuse

sin(θ) = -15/17

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The cosine ratio will involve the adjacent and hypotenuse

cos(angle) = adjacent/hypotenuse

cos(θ) = 8/17

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Tangent is opposite over adjacent

tan(angle) = opposite/adjacent

tan(θ) = -15/8

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The functions csc(θ), sec(θ), and cot(θ) are the functions cosecant, secant, and cotangent in that order.

They represent the reciprocal of sine, cosine, and tangent in that order as well.

Eg: sec(θ) is the reciprocal of cos(θ)

This means,

csc(θ) = -17/15 which is the flip of sin(θ) = -15/17
sec(θ) = 17/8 which is the flip of cos(θ) = 8/17
cot(θ) = -8/15 which is the flip of tan(θ) = -15/8

As an alternative, you could also say

csc(θ) = hypotenuse/opposite = 17/(-15) = -17/15
sec(θ) = hypotenuse/adjacent = 17/8
cot(θ) = adjacent/opposite = -15/8

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Here's a list of formulas that you can write down on a reference sheet

sin(θ) = y/h
cos(θ) = x/h
tan(θ) = y/x
csc(θ) = h/y
sec(θ) = h/x
cot(θ) = x/y

where h is the hypotenuse and h = sqrt(x^2+y^2). Notice that something like tan = y/x is the reciprocal of cot = x/y. The x and y refers to the coordinates of the terminal point given to you. Make sure to reduce all fractions as much as possible. In our case, none of the fractions can be reduced any further.