Okay so I have these vectors of magnitude 5.3 units and the cos theta between them is 5/13. So i thought of using the formula for vector addition: Square root of [ P^2 + Q^2 + 2PQcostheta ]. But i got the answer wrong and when i checked the textbook, they had done it like this:
Vector[P+Q] = Pcostheta + Qcostheta
What formula is this and why is mine wrong?
Vector[P+Q] = Pcostheta + Qcostheta
What formula is this and why is mine wrong?

You are right and the book is wrong but if their theta is half of cos^1(5/13) an d magnitudes are equal then both are right and final answer must be same.
This is how it can be proved that both could be right!
R = sq rt[P^2 + P^2 +2*(P^2)cos(theta)]
= sq rt[2*(P^2){1+cos(theta)}} =sq rt[2*(P^2){1+2*cos^2(theta/2)1}} = 2*P*cos(theta/2)
= P*cos(theta/2) + Q*cos(theta/2), with P = Q
When magnitudes are equal Resultant is symmetrically situated and the sin components cancel!
This is how it can be proved that both could be right!
R = sq rt[P^2 + P^2 +2*(P^2)cos(theta)]
= sq rt[2*(P^2){1+cos(theta)}} =sq rt[2*(P^2){1+2*cos^2(theta/2)1}} = 2*P*cos(theta/2)
= P*cos(theta/2) + Q*cos(theta/2), with P = Q
When magnitudes are equal Resultant is symmetrically situated and the sin components cancel!