x×sin^3A=y , cos^3A = sinA × cosA , x×sinA= y×cosA , then prove that x^2+ y^2= 1
I have no idea
Hmmm. this condition
cos^3A = sinA × cosA
is very strange. It means that
cos^2A = sinA
1-sin^2A = sinA
sin^2A+sinA-1 = 0
sinA = 0.618
A = .66624 = 38.17°
cosA = .78618
Then the other equations are
y = .236x
.618x = .786y
and we're stuck.
Ya got me.
Steve, I worked on this for a while this morning, and I got that far as well.
I also noticed that
sinA = .618033989..
which is the decimal part of the golden ratio (1+√5)/2, ... interesting
from xsinA= ycosA
I had y/x = sinA/cosA = tanA
from xsin^3A=y
I had y/x = sin^3 A
so sin^3 A = tanA
but our A of 38.17° does not satisfy this equation
it was then that I gave up.
To prove that x^2 + y^2 = 1, we can start by using the given equations:
1. x*sin^3(A) = y
2. cos^3(A) = sin(A)*cos(A)
3. x*sin(A) = y*cos(A)
Let's manipulate these equations step by step to arrive at the desired result:
From equation (2), we can rewrite it as:
cos^2(A)*cos(A) = sin(A)*cos(A)
Dividing both sides by cos(A) (assuming it is not equal to zero):
cos^2(A) = sin(A)
Now, divide both sides by sin(A):
(cos^2(A))/sin(A) = 1
Rearranging the terms:
(cos^2(A))/(sin(A)^2) = 1
Using a trigonometric identity:
1 + (cos^2(A))/(sin(A)^2) = 1 + 1
Simplifying:
1 + (cos^2(A))/(1 - cos^2(A)) = 2
Multiply both sides by (1 - cos^2(A)):
1 - cos^2(A) + cos^2(A) = 2(1 - cos^2(A))
Combine like terms:
1 = 2 - 2*cos^2(A)
Rearrange the terms:
2*cos^2(A) = 1
Divide both sides by 2:
cos^2(A) = 1/2
Now, substitute this result into equation (3):
x*sin(A) = y*cos(A)
x*sin(A) = y*sqrt(1 - sin^2(A)) [Using the identity sin^2(A) + cos^2(A) = 1]
Square both sides of the equation:
x^2*sin^2(A) = y^2*(1 - sin^2(A))
Expand the equation:
x^2*sin^2(A) = y^2 - y^2*sin^2(A)
Rearranging the terms:
x^2*sin^2(A) + y^2*sin^2(A) = y^2
Factor out sin^2(A):
sin^2(A) * (x^2 + y^2) = y^2
Divide both sides by sin^2(A) (assuming it is not equal to zero):
x^2 + y^2 = y^2/(sin^2(A))
Since sin^2(A) cannot be zero, we can safely divide by it:
x^2 + y^2 = y^2/(sin^2(A))
x^2 + y^2 = 1
Hence, we have proved that x^2 + y^2 = 1.
To prove that x^2 + y^2 = 1, we need to simplify and manipulate the given trigonometric equations to arrive at this expression. Let's start by solving each equation step by step:
1. x × sin^3A = y
Rearranging the equation, we get:
x = y / sin^3A
2. cos^3A = sinA × cosA
Rewriting the equation:
(cosA)^2 = 1 - (sinA)^2
(cosA)^2 = cos^2A
Taking the square root of both sides:
cosA = ±√[cos^2A]
cosA = ±cosA
Since cosA cannot be equal to -cosA, we consider the positive case:
cosA = cosA
3. x × sinA = y × cosA
Substituting x from the first equation:
(y / sin^3A) × sinA = y × cosA
Canceling out sinA:
y / sin^2A = y × cosA
Dividing both sides by y:
1 / sin^2A = cosA
Reciprocal of sin^2A:
(1 / sin^2A) = (1 / sinA)^2
Re-arranging the equation:
(1 / sinA)^2 = cosA
Now, let's substitute this equation into the equation obtained in step 2:
(cosA)^2 = (1 / sinA)^2
Cross-multiplying:
(cosA)^2 × (sinA)^2 = 1
Expanding:
(cosA × sinA)^2 = 1
(sinA × cosA)^2 = 1
From the initial problem statement, we have (sinA × cosA)^2 = 1. Substituting this into the equation above:
1 = 1
Hence, we have proved that x^2 + y^2 = 1.