Question 893601
The Maclaurin series expansion is,
{{{f(x)=sum(f^(k)(0)*((x-0)^k/k!),k=0,infinity)}}}
where {{{f^k}}} is the k-th derivative of the function.
In this case,
{{{f(x)=sin(x)}}}
{{{f^(1)(x)=cos(x)}}}
{{{f^(2)(x)=-sin(x)}}}
{{{f^(3)(x)=-cos(x)}}}
{{{f^(4)(x)=sin(x)}}}
{{{f^(5)(x)=cos(x)}}}
{{{f^(6)(x)=-sin(x)}}}
Then take the value of each at {{{x=0}}}
{{{f(0)=sin(0)=0}}}
{{{f^(1)(0)=cos(0)=1}}}
{{{f^(2)(0)=-sin(0)=0}}}
{{{f^(3)(0)=-cos(0)=-1}}}
{{{f^(4)(0)=sin(0)=0}}}
{{{f^(5)(0)=cos(0)=1}}}
{{{f^(6)(0)=-sin(0)=0}}}
Putting it all together,
{{{sin(x)=0+1((x-0))/1+0-1((x-0)^3)/3!-0+1*(x-0)^5/5!-0}}}
You begin to see the pattern,
{{{sin(x)=x-(x)^3/3!+(x)^5/5!-(x)^7/7!}}}
So find the value of 45 degrees in radians,
{{{45((2*pi)/360)=0.785398}}}
And substitute,
{{{sin(0.785398)=0.785398-(0.785398)^3/3!+(0.785398)^5/5!-(0.785398)^7/7!}}}
{{{sin(0.785398)=0.707106}}}
I think that's pretty good compared to {{{sqrt(2)/2=0.707107}}} 
{{{sin(0.785398)=0.7071}}}