Question

Expand sin x in ascending powers `x - pi/4` upto three non-zero terms

Answer 1

Let f(x) = sin x

f(x) = sin x, `"f"(pi/4) = 1/sqrt(2)`

f(x) = cos x, `"f'"(pi/4) = 1/sqrt(2)`

f(x) = – sin x, `"f''"(pi/4) = - 1/sqrt(2)`

f(x) = – cos x, `"f'''"(pi/4) = - 1/sqrt(2)`

Taylor series of f(x) is

f(x) = `sum_("n" = 0)^("n" = oo) "a"_"n" (x - pi/4)^pi`

Where a_n = `("f"^(("n"))(pi/4))/("n"!)`

∴ The required expansion is

sin x = `1/sqrt(2) + 1/sqrt(2) ((x - pi/4))/(1!) - 1/sqrt(2) ((x - pi/4)^2)/(2!) - 1/sqrt(2) (x - pi/4)^3/(3!) + ...`

= `1/sqrt(2) [1 + ((x - pi/4))/(1!) - (x - pi/4)^2/(2!) - (x - pi/4)^3/(3!) + ...]`

= `sqrt(2)/2 [1 + ((x - pi/4))/(1!) - (x - pi/4)^2/(2!) - (x - pi/4)^2/(3!) + ...]`