The period of cot(x/4)+tan(x/4)\1+tan(x/2)-tanx?
Period of the Function:
To find the period of the given function, we need to determine the smallest positive value of x for which the function repeats its values.
Expressing the Function:
Let's start by expressing the given function:
f(x) = (cot(x/4) * tan(x/4)) / (1 - tan(x/2) - tan(x))
Simplifying the Function:
To simplify the function, we can use trigonometric identities:
1. cot(x) = 1 / tan(x)
2. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
Applying these identities, we can simplify the function as follows:
f(x) = (cot(x/4) * tan(x/4)) / (1 - tan(x/2) - tan(x))
= ((1 / tan(x/4)) * tan(x/4)) / (1 - tan(x/2) - tan(x))
= 1 / (1 - tan(x/2) - tan(x))
Finding the Period:
To find the period of the function, we need to find the value of x for which the function repeats its values. In other words, we need to find the smallest positive value of x such that f(x) = f(x + T), where T represents the period of the function.
Let's substitute x = x + T into the function:
f(x + T) = 1 / (1 - tan((x + T)/2) - tan(x + T))
Since we want f(x + T) to be equal to f(x), we can set the two expressions equal to each other:
1 / (1 - tan(x/2) - tan(x)) = 1 / (1 - tan((x + T)/2) - tan(x + T))
Key Point:
The key point to notice here is that the denominator of the function is the same in both expressions.
Analyzing the Denominator:
Let's simplify the denominator of the function:
1 - tan(x/2) - tan(x) = 1 - tan(x/2) - [tan(x/2 + x/2)]
= 1 - tan(x/2) - [(tan(x/2) + tan(x/2)) / (1 - tan(x/2)tan(x/2))]
= 1 - tan(x/2) - 2tan(x/2) / (1 - tan(x/2)tan(x/2))
= (1 - 2tan(x/2)) / (1 - tan(x/2)tan(x/2))
Setting the Expressions Equal:
Now, let's set the two expressions equal to each other:
1 / (1 - tan(x/2) - tan(x)) = 1 / (1 - tan((x + T)/2) - tan(x + T))
=> (1 - 2tan(x/2)) / (1 - tan(x/2)tan(x/2)) = (1 - 2tan((x + T)/2)) / (1 - tan((x + T)/2)tan((x + T)/2))
Key Point:
The key point to notice here is that
The period of cot(x/4)+tan(x/4)\1+tan(x/2)-tanx?
4π