Question

Expansion of sin(x/2)+cos(x/2) in ascending power of x
​

Answer 1

Answer:

1+(x/2)-(x^2/8)-(x^3/48)

Step-by-step explanation:

Answer 2

Answer:

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

Step-by-step explanation:

Using TAYLOR's theoram about x=0, we have

sin($\frac{x}{2}$)=$\frac{x}{2}$ - $\frac{x^{3} }{3!(2^{3} )}$ + $\frac{x^{5} }{5!(2^{5} )}$ - $\frac{x^{7} }{7!(2^{7}) }$+....             eq.(1)

cos($\frac{x}{2}$)=1 - $\frac{x^{2} }{2!(2^{2} )}$ + $\frac{x^{4} }{4!(2^{4} )}$ - $\frac{x^{6} }{6!(2^{6} )}$+.....           eq.(2)

adding eq 1+2,we get.

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