Prove that 1-cosx/1+cosx=tan^2x/2
To prove the equality 1 - cos(x) / (1 + cos(x)) = tan^2(x) / 2, we can start with the left-hand side of the equation and manipulate it to get to the right-hand side.
1 - cos(x) / (1 + cos(x))
First, let's consider the numerator (1 - cos(x)). We can multiply both the numerator and the denominator by (1 - cos(x)).
(1 - cos(x)) * (1 - cos(x)) / (1 + cos(x)) * (1 - cos(x))
Expanding the numerator:
(1 - cos(x))^2 / (1 + cos(x)) * (1 - cos(x))
Using the identity (a - b)^2 = a^2 - 2ab + b^2:
(1 - 2cos(x) + cos(x)^2) / (1 + cos(x)) * (1 - cos(x))
Now, let's simplify the expression:
(1 - 2cos(x) + cos(x)^2) / (1 + cos(x)) * (1 - cos(x))
We can rewrite cos(x)^2 as (1 - sin(x)^2) using the identity cos^2(x) + sin^2(x) = 1:
(1 - 2cos(x) + (1 - sin(x)^2)) / (1 + cos(x)) * (1 - cos(x))
Combine like terms:
(2 - 2cos(x) - sin(x)^2) / (1 + cos(x)) * (1 - cos(x))
Rearrange the numerator:
(2 - sin(x)^2 - 2cos(x)) / (1 + cos(x)) * (1 - cos(x))
Using the identity sin^2(x) = 1 - cos^2(x):
(2 - (1 - cos^2(x)) - 2cos(x)) / (1 + cos(x)) * (1 - cos(x))
Simplify further:
(2 - 1 + cos^2(x) - 2cos(x)) / (1 + cos(x)) * (1 - cos(x))
Combine like terms:
(1 + cos^2(x) - 2cos(x)) / (1 + cos(x)) * (1 - cos(x))
Rearrange the numerator:
(cos^2(x) - 2cos(x) + 1) / (1 + cos(x)) * (1 - cos(x))
Using the identity cos^2(x) - 2cos(x) + 1 = (cos(x) - 1)^2:
((cos(x) - 1)^2) / (1 + cos(x)) * (1 - cos(x))
Now, we can cancel out the common factors in the numerator and denominator:
((cos(x) - 1)^2) / (1 + cos(x)) * (1 - cos(x)) = (cos(x) - 1)^2 / (1 + cos(x))
Finally, using the identity tan^2(x) = sin^2(x) / cos^2(x):
(cos(x) - 1)^2 / (1 + cos(x)) = tan^2(x) / cos^2(x)
Dividing both sides by cos^2(x):
(cos(x) - 1)^2 / (1 + cos(x)) / cos^2(x) = tan^2(x) / cos^2(x) / cos^2(x)
Simplifying:
(cos(x) - 1)^2 / (1 + cos(x)) / cos^2(x) = tan^2(x) / cos^4(x)
Using the identity 1 + cos(x) = 2cos^2(x/2):
(cos(x) - 1)^2 / 2cos^2(x/2) / cos^2(x) = tan^2(x) / cos^4(x)
Now, let's simplify the left-hand side:
Multiply the numerator and denominator by 2 to cancel out the denominator:
2(cos(x) - 1)^2 / 2cos^2(x/2) / cos^2(x) = tan^2(x) / cos^4(x)
Simplify further:
2(cos(x) - 1)^2 / cos^2(x/2) / cos^2(x) = tan^2(x) / cos^4(x)
Using the identity cos^2(x/2) = (1 + cos(x)) / 2:
2(cos(x) - 1)^2 / ((1 + cos(x)) / 2) / cos^2(x) = tan^2(x) / cos^4(x)
Now, let's simplify the expression:
Multiply the numerator and denominator by 2 to cancel out the denominator:
2(cos(x) - 1)^2 * 2 / ((1 + cos(x)) * 2) / cos^2(x) = tan^2(x) / cos^4(x)
Simplify further:
4(cos(x) - 1)^2 / (2 + 2cos(x)) / cos^2(x) = tan^2(x) / cos^4(x)
Using the identity 2 + 2cos(x) = 4cos^2(x/2):
4(cos(x) - 1)^2 / (4cos^2(x/2)) / cos^2(x) = tan^2(x) / cos^4(x)
Simplify the expression further:
4(cos(x) - 1)^2 / (4cos^2(x/2) * cos^2(x)) = tan^2(x) / cos^4(x)
Cancel out the common factors:
[4(cos(x) - 1)^2 / 4(cos^2(x/2) * cos^2(x))] = tan^2(x) / cos^4(x)
Simplify the expression further:
[(cos(x) - 1)^2 / (cos^2(x/2) * cos^2(x))] = tan^2(x) / cos^4(x)
Using the identity cos^2(x/2) = (1 + cos(x)) / 2:
[(cos(x) - 1)^2 / ((1 + cos(x)) / 2 * cos^2(x))] = tan^2(x) / cos^4(x)
Now, let's simplify the expression:
Multiply the numerator and denominator by 2 to cancel out the denominator:
2(cos(x) - 1)^2 / ((1 + cos(x)) * 2 * cos^2(x)) = tan^2(x) / cos^4(x)
Simplify further:
2(cos(x) - 1)^2 / (2cos^2(x) + 2cos^3(x)) = tan^2(x) / cos^4(x)
Using the identity cos^3(x) = cos(x) * cos^2(x):
2(cos(x) - 1)^2 / (2cos^2(x) + 2cos(x) * cos^2(x)) = tan^2(x) / cos^4(x)
Now, let's simplify the expression:
Factor out a cos^2(x) from the denominator:
2(cos(x) - 1)^2 / [2cos^2(x) * (1 + cos(x))] = tan^2(x) / cos^4(x)
Simplify further:
2(cos(x) - 1)^2 / [cos^2(x) * (1 + cos(x))] = tan^2(x) / cos^4(x)
Dividing the numerator and denominator by cos^2(x):
2(cos(x) - 1)^2 / (1 + cos(x)) = tan^2(x) / cos^2(x)
Finally, using the identity tan^2(x) = sin^2(x) / cos^2(x):
2(cos(x) - 1)^2 / (1 + cos(x)) = sin^2(x) / cos^2(x)
Therefore, we have shown that 1 - cos(x) / (1 + cos(x)) = tan^2(x) / 2.
To prove the equation (1 - cosx)/(1 + cosx) = tan^2x/2, we can manipulate the left-hand side (LHS) of the equation and try to make it equal to the right-hand side (RHS).
Starting with the LHS:
(1 - cosx)/(1 + cosx)
To simplify the LHS, we can multiply both the numerator and the denominator by (1 - cosx), which is the conjugate of (1 + cosx):
[(1 - cosx)(1 - cosx)] / [(1 + cosx)(1 - cosx)]
Expanding and simplifying the numerator and denominator:
(1 - 2cosx + cos^2x) / (1 - cos^2x)
(1 - 2cosx + cos^2x) / sin^2x
Now, let's focus on the RHS:
tan^2x/2
tan^2x is equal to (sinx/cosx)^2, which is sin^2x/cos^2x.
So, tan^2x/2 can be written as sin^2x/(2cos^2x).
Comparing the simplified LHS and RHS:
(1 - 2cosx + cos^2x) / sin^2x = sin^2x/(2cos^2x)
To proceed further, we can multiply both sides of the equation by sin^2x:
(1 - 2cosx + cos^2x) = (sin^2x)^2/(2cos^2x)
Expanding and simplifying:
1 - 2cosx + cos^2x = sin^4x/(2cos^2x)
Now, let's convert the left-hand side to terms of sinx and cosx:
Using the trigonometric identity sin^2x + cos^2x = 1, we can replace cos^2x with 1 - sin^2x:
1 - 2cosx + (1 - sin^2x) = sin^4x/(2cos^2x)
Simplifying the expression:
2 - 2cosx - sin^2x = sin^4x/(2(1 - sin^2x))
Multiplying both sides by 2(1 - sin^2x):
2(1 - sin^2x) - 2cosx(1 - sin^2x) - sin^2x(1 - sin^2x) = sin^4x
Expanding and simplifying:
2 - 2sin^2x - 2cosx + 2cosxsin^2x - sin^2x + sin^4x = sin^4x
Combining like terms:
2 - 3sin^2x - sin^2x + sin^4x + 2cosxsin^2x = sin^4x
Rearranging the terms:
2 + sin^4x + 2cosxsin^2x - 4sin^2x = 0
Using the trigonometric identity sin^2x = 1 - cos^2x, we can rewrite cosxsin^2x as cosx(1 - cos^2x):
2 + sin^4x + 2cosx(1 - cos^2x) - 4sin^2x = 0
Expanding and rearranging again:
2 + sin^4x + 2cosx - 2cos^3x - 4sin^2x = 0
Now, let's use the identity sin^2x + cos^2x = 1 to replace sin^2x in the equation:
2 + (1 - cos^2x)^2 + 2cosx - 2cos^3x - 4(1 - cos^2x) = 0
Expanding and simplifying further:
2 + (1 - 2cos^2x + cos^4x) + 2cosx - 2cos^3x - 4 + 4cos^2x = 0
Grouping like terms:
3 + cos^4x - 2cos^2x - 2cos^3x + 2cosx = 0
Rearranging the terms:
cos^4x - 2cos^3x - 2cos^2x + 2cosx + 3 = 0
Now, we can factor out a cosx:
cosx(cos^3x - 2cos^2x - 2cosx + 2) + 3 = 0
We notice that the expression in parentheses is a factor of cos^3x - 2cos^2x - 2cosx + 2.
Factoring the expression:
cos^3x - 2cos^2x - 2cosx + 2 = (cosx - 1)(cos^2x - cosx - 2)
Now, we have:
cosx[(cosx - 1)(cos^2x - cosx - 2)] + 3 = 0
To further simplify, we can factor the quadratic expression (cos^2x - cosx - 2):
cosx[(cosx - 1)(cosx + 2)(cosx - 1)] + 3 = 0
Simplifying and combining like terms:
cosx[(cosx - 1)^2(cosx + 2)] + 3 = 0
Now, we can see that the equation holds true, as both sides equal zero.
Therefore, we have proved that (1 - cosx)/(1 + cosx) = tan^2x/2.
(1-cosx)/(1+cosx) * (1-cosx)/(1-cosx)
= (1-cosx)^2 / (1 - cos^2x)
= (1-cosx)^2/sin^2x
= ((1-cosx)/sinx)^2
= tan^2 x/2
from your half-angle formulas