please answer this early
Attachments:
Asahithi:
urgent
Answers
Answered by
1
1 = (cos(A))^4 / (cos(B))^2 + (sin(A))^4 / (sin(B))^2
define:
a = cos(A)
b = cos(B)
Notice that
sin(A) = (1 - (cos(A))^2) = sqrt(1-a^2) , so:
(sin(A))^2 = 1 - a^2 , and likewise:
(sin(B))^2 = 1 - b^2
Therefore, your original equation becomes:
1 = a^4/b^2 + (1 - a^2)^2/(1 - b^2)
Multiply both sides by (b^2) * (1 - b^2):
(b^2) * (1 - b^2) = (a^4)*(1 - b^2) + (b^2)*(1 - a^2)^2
b^2 - b^4 = a^4 - (a^4)*(b^2) + (b^2)*(1 - 2*a^2 + a^4)
b^2 - b^4 = a^4 - (a^4)*(b^2) + b^2 - 2*(a^2)*(b^2) + (a^4)(b^2)
Cancelling out some terms:
- b^4 = a^4 - 2*(a^2)(b^2) , or
0 = a^4 - 2*(a^2)(b^2) + b^4
= (a^2 - b^2)^2
=> a^2 = b^2
OK, that means
(cos(B))^4 = b^4
(cos(A))^2 = a^2 = b^2
so:
(cos(B))^4 / (cos(A))^2 = b^4 / b^2 = b^2
It also means that:
(sin(B))^4 = ((sin(B))^2)^2 = (1 - b^2)^2 , and
(sin(A))^2 = (1 - a^2) = (1 - b^2)
so:
(sin(B))^4 / (sin(A))^2 = (1 - b^2)^2 / (1 - b^2) = 1 - b^2
So when you add the sum together, you get:
(cos(B))^4 / (cos(A))^2 + (sin(B))^4 / (sin(A))^2 = b^2 + 1 - b^2 = 1
define:
a = cos(A)
b = cos(B)
Notice that
sin(A) = (1 - (cos(A))^2) = sqrt(1-a^2) , so:
(sin(A))^2 = 1 - a^2 , and likewise:
(sin(B))^2 = 1 - b^2
Therefore, your original equation becomes:
1 = a^4/b^2 + (1 - a^2)^2/(1 - b^2)
Multiply both sides by (b^2) * (1 - b^2):
(b^2) * (1 - b^2) = (a^4)*(1 - b^2) + (b^2)*(1 - a^2)^2
b^2 - b^4 = a^4 - (a^4)*(b^2) + (b^2)*(1 - 2*a^2 + a^4)
b^2 - b^4 = a^4 - (a^4)*(b^2) + b^2 - 2*(a^2)*(b^2) + (a^4)(b^2)
Cancelling out some terms:
- b^4 = a^4 - 2*(a^2)(b^2) , or
0 = a^4 - 2*(a^2)(b^2) + b^4
= (a^2 - b^2)^2
=> a^2 = b^2
OK, that means
(cos(B))^4 = b^4
(cos(A))^2 = a^2 = b^2
so:
(cos(B))^4 / (cos(A))^2 = b^4 / b^2 = b^2
It also means that:
(sin(B))^4 = ((sin(B))^2)^2 = (1 - b^2)^2 , and
(sin(A))^2 = (1 - a^2) = (1 - b^2)
so:
(sin(B))^4 / (sin(A))^2 = (1 - b^2)^2 / (1 - b^2) = 1 - b^2
So when you add the sum together, you get:
(cos(B))^4 / (cos(A))^2 + (sin(B))^4 / (sin(A))^2 = b^2 + 1 - b^2 = 1
Similar questions