1 - (sin^2x / 1+cosx) = cosx
1 - ( [1-cos^2]/[1+cos] ) = ? cos
1 - (1-cos)(1+cos)/(1+cos) = ? cos
1 - (1-cos) = ? cos
yes
cos = cos :)
To solve the equation 1 - (sin^2x / 1+cosx) = cosx, we need to simplify and manipulate the equation to isolate the variable x.
Let's start by simplifying the left side of the equation:
1 - (sin^2x / 1+cosx) = cosx
Since sin^2x is the same as (sinx)^2, we can rewrite the equation as:
1 - sin^2x / (1 + cosx) = cosx
Next, since (1 + cosx) is in the denominator, we can multiply both sides of the equation by the denominator to eliminate the fraction:
(1 + cosx) * (1 - sin^2x / (1 + cosx)) = cosx * (1 + cosx)
Simplifying further:
(1 + cosx) - sin^2x = cosx * (1 + cosx)
Now, let's simplify the equation step by step:
1 + cosx - sin^2x = cosx + cos^2x
Rearranging the equation:
1 - sin^2x = cos^2x
Using the trigonometric identity sin^2x + cos^2x = 1:
1 - sin^2x = 1 - cos^2x
Since both sides of the equation are equal to 1, we can cancel the 1:
-sin^2x = -cos^2x
Dividing both sides of the equation by -1:
sin^2x = cos^2x
Now, take the square root of both sides of the equation:
sinx = ±cosx
To solve for x, we can set up two equations:
1. sinx = cosx
In this case, we can divide both sides of the equation by cosx:
tanx = 1
Taking the inverse tangent of both sides:
x = arctan(1) + kπ, where k is an integer
2. sinx = -cosx
In this case, we can divide both sides of the equation by cosx:
tanx = -1
Taking the inverse tangent of both sides:
x = arctan(-1) + kπ, where k is an integer
Therefore, the solutions for x are given by:
x = arctan(1) + kπ,
x = arctan(-1) + kπ
where k is an integer value that represents the period of the trigonometric functions.
To simplify the left-hand side of the equation, we'll start by working with the denominator.
1 + cos(x) can be written as cos(x) + 1.
Now, let's substitute this back into the equation:
1 - (sin^2x / (cos(x) + 1)) = cos(x)
Next, let's simplify the left-hand side by using the fact that sin^2(x) = 1 - cos^2(x):
1 - ((1 - cos^2(x)) / (cos(x) + 1)) = cos(x)
Now, let's combine like terms on the left-hand side:
(1 * (cos(x) + 1) - (1 - cos^2(x))) / (cos(x) + 1) = cos(x)
Simplifying further:
(cos(x) + 1 - 1 + cos^2(x)) / (cos(x) + 1) = cos(x)
(cos(x) + cos^2(x)) / (cos(x) + 1) = cos(x)
Now, let's simplify the numerator by factoring out cos(x):
cos(x) * (1 + cos(x)) / (cos(x) + 1) = cos(x)
We can cancel out the common factors of (cos(x) + 1) on both the numerator and denominator:
cos(x) = cos(x)
Therefore, the original equation 1 - (sin^2x / (1 + cosx)) = cosx is true for all values of x.