Expand cos x cos y in powers of x and y.
Answers
Step-by-step explanation:
if x = 1/(square root(2) -1 ) then prove that x^2 - 6 + 1/x^2 = 0
Answer:
The expansion of cos x cos y in powers of x and y gives us an approximate value cos x cos y = 1 - x^2/2 - y^2/2 + x^4/24 + x^2y^2/4 + y^4/24 + ...
Step-by-step explanation:
The expansion of cos x cos y in powers of x and y can be obtained using the Taylor series expansion of cosine.
The Taylor series expansion is a powerful tool for finding the approximate value of a function. The idea behind this method is to represent a function as a power series of its derivatives at a certain point.
In the case of cos x cos y, we want to find the expansion of this function in powers of x and y. To do this, we start by finding the Taylor series expansion of cos x and cos y individually.
The Taylor series expansion of cos x is given by:
cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ...
Here, x^2/2!, x^4/4!, etc., are the derivatives of cos x at the point x=0.
Similarly, the Taylor series expansion of cos y is given by:
cos y = 1 - y^2/2! + y^4/4! - y^6/6! + ...
Next, we multiply these two expansions to obtain the expansion of cos x cos y:
cos x cos y = (1 - x^2/2! + x^4/4! - x^6/6! + ...)(1 - y^2/2! + y^4/4! - y^6/6! + ...)
Expanding the product, we get:
cos x cos y = 1 - (x^2 + y^2)/2! + (x^2 y^2 - x^4/2! - y^4/2!)/4! + ...
Finally, we simplify the expansion by dividing each term by the corresponding factorial:
cos x cos y = 1 - x^2/2 - y^2/2 + x^4/24 + x^2y^2/4 + y^4/24 + ...
This is the desired result. The expansion of cos x cos y in powers of x and y gives us an approximate value of the function for small values of x and y. For larger values of x and y, higher order terms in the expansion may become important.
To learn more about derivatives from the link below
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