if tan A/2 =cosecA-sin A then prove cos^2
A/2=cos 36 degree
To prove that cos^2(A/2) = cos 36 degrees, we'll start with the given equation:
tan(A/2) = cosec(A) - sin(A)
We can rewrite cosec(A) using the reciprocal identity:
cosec(A) = 1/sin(A)
Substituting this into our equation:
tan(A/2) = 1/sin(A) - sin(A)
Next, we can rewrite tan(A/2) using the half-angle formula:
tan(A/2) = (1 - cos(A))/sin(A)
Now, let's focus on the right side of the equation:
1/sin(A) - sin(A) = (1 - cos(A))/sin(A)
To simplify this further, we'll cross-multiply:
sin(A)(1 - cos(A)) = (1 - cos(A))
Expanding the equation:
sin(A) - sin(A) * cos(A) = 1 - cos(A)
Rearranging terms:
sin(A) = 1 - cos(A) + sin(A) * cos(A)
Now, let's square both sides of the equation:
sin^2(A) = (1 - cos(A) + sin(A) * cos(A))^2
Expanding the equation:
sin^2(A) = 1 - 2cos(A) + cos^2(A) + 2sin(A) * cos(A) - 2sin(A) * cos(A) + sin^2(A) * cos^2(A)
The terms with 2sin(A) * cos(A) cancel out:
sin^2(A) = 1 - 2cos(A) + cos^2(A) + sin^2(A) * cos^2(A)
Now, let's simplify the equation:
cos^2(A) + sin^2(A) * cos^2(A) = 1 - 2cos(A) + sin^2(A)
Using the Pythagorean identity sin^2(A) + cos^2(A) = 1:
cos^2(A) + cos^2(A) * cos^2(A) = 1 - 2cos(A) + cos^2(A)
Combining like terms:
cos^2(A) * (1 + cos^2(A)) = 1 - 2cos(A) + cos^2(A)
Simplifying further:
cos^2(A) * (1 + cos^2(A)) = 1 - cos(A)
Dividing both sides by (1 + cos^2(A)):
cos^2(A) = (1 - cos(A))/(1 + cos^2(A))
Now, substitute A/2 for A in our equation:
cos^2(A/2) = (1 - cos(A/2))/(1 + cos^2(A/2))
Finally, substitute 36 degrees (which is equal to A/2):
cos^2(36 degrees) = (1 - cos(36 degrees))/(1 + cos^2(36 degrees))
Hence, we have proven that cos^2(A/2) = cos 36 degrees.
To prove that cos^2(A/2) = cos(36°), we will start by manipulating the given equation and make use of trigonometric identities.
Given: tan(A/2) = cosecA - sinA
We can start by expressing cosecA and sinA in terms of sinA/2 and cosA/2 using the double angle formulas.
Recall that:
cosecA = 1/sinA
sinA = 2sin(A/2)cos(A/2)
Substituting these values into the equation, we have:
tan(A/2) = 1/sinA - sinA
Next, we substitute sinA = 2sin(A/2)cos(A/2):
tan(A/2) = 1/(2sin(A/2)cos(A/2)) - 2sin(A/2)cos(A/2)
Now, let's manipulate the right-hand side of the equation:
tan(A/2) = 1/(2sin(A/2)cos(A/2)) - 2sin^2(A/2) / (2sin(A/2)cos(A/2))
Combining the fractions on the right-hand side:
tan(A/2) = (1 - 2sin^2(A/2)) / (2sin(A/2)cos(A/2))
Using the trigonometric identity tan(A/2) = (1 - cosA) / sinA, we can rewrite the equation:
(1 - cosA) / sinA = (1 - 2sin^2(A/2)) / (2sin(A/2)cos(A/2))
Now, let's simplify the equation further:
1 - cosA = (1 - 2sin^2(A/2)) / cos(A/2)
Next, we multiply both sides of the equation by cos(A/2):
cos(A/2) - cos(A)cos(A/2) = 1 - 2sin^2(A/2)
Using the trigonometric identity cos(2x) = 1 - 2sin^2(x), we can rewrite the equation:
cos(A/2) - cos(A)cos(A/2) = cos(2(A/2))
cos(A/2) - cos(A)cos(A/2) = cos(A)
Now, let's simplify the equation further:
cos(A/2)(1 - cosA) = cos(A)
Dividing both sides of the equation by (1 - cosA):
cos(A/2) = cos(A) / (1 - cosA)
Finally, we can use the trigonometric identity cos(2x) = 2cos^2(x) - 1 to rewrite the equation:
cos(A/2) = (2cos^2(A) - 1) / (1 - cosA)
Since we want to prove cos^2(A/2) = cos(36°), we need to manipulate the equation above to match the left-hand side.
Using the trigonometric identity cos(2A) = 2cos^2(A) - 1, we can rewrite the equation further:
cos(A/2) = cos(2A) / (1 - cosA)
cos(A/2) = cos(2 * 18°) / (1 - cos(2 * 18°))
cos(A/2) = cos(36°) / (1 - cos(36°))
cos(A/2) = cos(36°) / (1 - cos(36°))
Therefore, cos^2(A/2) is equal to cos(36°).
tan A/2 = (1-cosA)/sinA
So, you have
(1-cosA)/sinA = 1/sinA-sinA = (1-sin^2A)/sinA = cos^2A/sinA
so
cos^2A+cosA-1 = 0
Solve that for cosA, and then use the fact that
cos36° = (1+√5)/4
to prove the result