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dusya [7]
3 years ago
11

What is the inverse of y=^x3

Mathematics
1 answer:
Rashid [163]3 years ago
4 0

Answer:

Inverse of y=x^3 is f^-1(x) = ∛x

Step-by-step explanation:

We need to find the inverse of y=x^3

Step 1:

Interchange the variables:

x= y^3

Step 2: Now solve to find the value of y

=> y^3 = x

taking cube root on both sides of the equation

∛y^3 = ∛x

y=∛x

Step 3: Replace y with f^-1(x)

f^-1(x) = ∛x

So inverse of y=x^3 is f^-1(x) = ∛x

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and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

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cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

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cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

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cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

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sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
3 years ago
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