Question
If x = p sin theater and y = b cos theater , show that b square x square + a square y square = a square b .
Answers
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Starting with the given expressions for x and y:
x = p sin θ
y = b cos θ
We can square both of these expressions:
x^2 = p^2 sin^2 θ
y^2 = b^2 cos^2 θ
Next, we use the identity sin^2 θ + cos^2 θ = 1 to eliminate one of the trig functions. Since we have p and b in the expression we want to prove, we can use the Pythagorean identity:
sin^2 θ = 1 - cos^2 θ
Substituting this into the expression for x^2:
x^2 = p^2 (1 - cos^2 θ)
Expanding this out:
x^2 = p^2 - p^2 cos^2 θ
And similarly for y^2:
y^2 = b^2 - b^2 sin^2 θ
y^2 = b^2 - b^2 (1 - cos^2 θ)
y^2 = b^2 cos^2 θ
Now we can substitute both of these expressions into the left-hand side of the equation we want to prove:
b^2 x^2 + a^2 y^2
= b^2 (p^2 - p^2 cos^2 θ) + a^2 (b^2 cos^2 θ)
= p^2 b^2 - p^2 b^2 cos^2 θ + a^2 b^2 cos^2 θ
= (p^2 b^2 cos^2 θ + p^2 b^2 sin^2 θ) + (a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ)
= p^2 b^2 + a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ
Now we can use the same Pythagorean identity as before to simplify:
sin^2 θ = 1 - cos^2 θ
sin^2 θ + cos^2 θ = 1
p^2 sin^2 θ + p^2 cos^2 θ = p^2
Substituting this into the expression we just derived:
b^2 x^2 + a^2 y^2 = p^2 b^2 + a^2 b^2 - p^2 b^2 sin^2 θ
= p^2 b^2 (1 - sin^2 θ) + a^2 b^2
= p^2 b^2 cos^2 θ + a^2 b^2
But recall that we had this same expression for b^2 x^2 + a^2 y^2 earlier:
b^2 x^2 + a^2 y^2 = a^2 b^2 cos^2 θ + p^2 b^2
So we have shown that these two expressions are equal:
a^2 b^2 cos^2 θ + p^2 b^2 = p^2 b^2 cos^2 θ + a^2 b^2
Rearranging:
a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ = a^2 b^2 - p^2 b^2
cos^2 θ (a^2 b^2 - p^2 b^2) = a^2 b^2 - p^2 b^2
cos^2 θ = (a^2 b^2 - p^2 b^2) / (a^2 b^2 - p^2 b^2)
cos^2 θ = 1
Therefore, the original equation
b^2 x^2 + a^2 y^2 = a^2 b^2
is true for any value of θ.
x = p sin θ
y = b cos θ
We can square both of these expressions:
x^2 = p^2 sin^2 θ
y^2 = b^2 cos^2 θ
Next, we use the identity sin^2 θ + cos^2 θ = 1 to eliminate one of the trig functions. Since we have p and b in the expression we want to prove, we can use the Pythagorean identity:
sin^2 θ = 1 - cos^2 θ
Substituting this into the expression for x^2:
x^2 = p^2 (1 - cos^2 θ)
Expanding this out:
x^2 = p^2 - p^2 cos^2 θ
And similarly for y^2:
y^2 = b^2 - b^2 sin^2 θ
y^2 = b^2 - b^2 (1 - cos^2 θ)
y^2 = b^2 cos^2 θ
Now we can substitute both of these expressions into the left-hand side of the equation we want to prove:
b^2 x^2 + a^2 y^2
= b^2 (p^2 - p^2 cos^2 θ) + a^2 (b^2 cos^2 θ)
= p^2 b^2 - p^2 b^2 cos^2 θ + a^2 b^2 cos^2 θ
= (p^2 b^2 cos^2 θ + p^2 b^2 sin^2 θ) + (a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ)
= p^2 b^2 + a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ
Now we can use the same Pythagorean identity as before to simplify:
sin^2 θ = 1 - cos^2 θ
sin^2 θ + cos^2 θ = 1
p^2 sin^2 θ + p^2 cos^2 θ = p^2
Substituting this into the expression we just derived:
b^2 x^2 + a^2 y^2 = p^2 b^2 + a^2 b^2 - p^2 b^2 sin^2 θ
= p^2 b^2 (1 - sin^2 θ) + a^2 b^2
= p^2 b^2 cos^2 θ + a^2 b^2
But recall that we had this same expression for b^2 x^2 + a^2 y^2 earlier:
b^2 x^2 + a^2 y^2 = a^2 b^2 cos^2 θ + p^2 b^2
So we have shown that these two expressions are equal:
a^2 b^2 cos^2 θ + p^2 b^2 = p^2 b^2 cos^2 θ + a^2 b^2
Rearranging:
a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ = a^2 b^2 - p^2 b^2
cos^2 θ (a^2 b^2 - p^2 b^2) = a^2 b^2 - p^2 b^2
cos^2 θ = (a^2 b^2 - p^2 b^2) / (a^2 b^2 - p^2 b^2)
cos^2 θ = 1
Therefore, the original equation
b^2 x^2 + a^2 y^2 = a^2 b^2
is true for any value of θ.