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BaLLatris [955]
3 years ago
11

A point on the terminal side of an angle theta in standard position is (negative 24 comma 18 ). Find the exact value of each of

the six trigonometric functions of theta.

Mathematics
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:

(I)Sin Θ=4/5

(II) Cos Θ =-3/5

(III) Tan Θ = -4/3

(IV) Cosec Θ = 5/4

(V) Sec Θ = -5/3

(VI) Cot Θ = -3/4

Step-by-step explanation:

(-24,18)

This lies in the second quadrant from the diagram.

Note that the angle is at the origin (0,0).

Using Pythagoras triples(18,24,30)

Hypotenuse = 30

Opposite=-24

Adjacent=18

(I)Sin Θ= Opposite/Hypotenuse

=-24/30=-4/5

Note: (Sine is Positive in the Second Quadrant)

Sin Θ =4/5

(II) Cos Θ = Adjacent/Hypotenuse

=18/30=3/5

Since Cosine is negative in the Second Quadrant

Cos Θ=-3/5

(III) Tan Θ = Opposite/Adjacent

Tan Θ = -24/18= -4/3

(IV) Cosec Θ = 1/Sin Θ = 5/4

(V) Sec Θ = 1/cos Θ = -5/4

(VI) Cot Θ = 1/tan Θ = -3/4

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Step-by-step explanation:

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18-12=6

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6 * 2= 12

find the area of the rectangle:

(18*4)-12=60

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3 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0
2 years ago
Find each function value<br><br>1. f(7) if (x)=5x<br><br>2. f(4) if f(x)=3x-1
Inessa [10]

<u>ANSWER</u>

1. \:  \: f(7) = 35

2. \: f(4) = 11

<u>EXPLANATION</u>

To find the value of a function, f(x) at a given value x=a, we plug in x=a into f(x) to obtain f(a) in simplified form.

The first function is

f(x) = 5x

To find f(7), we substitute x=7, into the function.

\implies f(7) =5(7) =  5 \times 7 = 35

The second function is

f(x) = 3x - 1

To find f(4), we replace x with 4 to obtain,

f(4) = 3(4) - 1

f(4) = 12 - 1 = 11

6 0
3 years ago
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