Question

expansion of sin x/2 + cos x/2 in ascending powers of x

Answer 1

Answer:

1

sin x /2 + cos x /2

(sin x + cos x ) /2

squaring on both numerator and denominator

sin x² + cos x² + 2 sin x. cos x whole divides by 2

Answer 2

Concept:

A series expansion, often known as an infinite sum, is the extension of a function into a series. It's a way of computing a function that can't be stated using basic operations like addition, subtraction, multiplication, and division. The resultant "series" may frequently be restricted to a finite number of terms, providing a function approximation.

Given:

The terms sin(x/2) and cos(x/2).

Find:

The expansion of the given terms.

Solution:

sin x = $x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\cdots$

cos x = $1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\cdots$

sin(x/2) = $\frac{x}{2} -\frac{x^{3}}{2^3*3 !}+\frac{x^{5}}{2^5*5 !}-\frac{x^{7}}{2^7*7 !}+\cdots$

cos(x/2) = $1-\frac{x^{2}}{2^2*2 !}+\frac{x^{4}}{2^4*4 !}-\frac{x^{6}}{2^6*6 !}+\cdots$

sin(x/2)+ cos(x/2) = $1+\frac{x}{2} -\frac{x^{2}}{2 !}-\frac{x^{3}}{2^3*3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{2^5*5 !}-\frac{x^{6}}{6 !}-\frac{x^{7}}{2^7*7 !}+\cdots$

Hence, the expansion of the terms is, $1+\frac{x}{2} -\frac{x^{2}}{2 !}-\frac{x^{3}}{2^3*3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{2^5*5 !}-\frac{x^{6}}{6 !}-\frac{x^{7}}{2^7*7 !}+\cdots$