Precalculus, Chapter 6, 6.3, Section 6.3, Problem 68

The magnitude of a vector u = a*i + b*j , such that:
|u| = sqrt(a^2+b^2)
Since the problem provides the magnitude |v| = 1 , yields:
1 = sqrt(a^2+b^2)
The direction angle of the vector can be found using the formula, such that:
tan theta = b/a
Since the problem provides the information that the direction angle of the vector v is theta = 45^o , yields:
tan 45^o= b/a => 1 = b/a => a = b
Replacing a for b in equation 1 = sqrt(a^2+b^2) yields:
1 = sqrt(a^2+a^2)=> 1 = +-a*sqrt 2 => a = +-(sqrt2)/2
b = +-(sqrt2)/2
Hence, the component form of the vector v can be <(sqrt2)/2,(sqrt2)/2> or <-(sqrt2)/2,-(sqrt2)/2>.