inv trig + complex qn (1 Viewer)

[Masaken]

i know in 3u you would solve this with the triangle technique, but how would you apply complex numbers here to prove the rhs is pi? thanks in advance, i'm so confused

 

[yanujw]

The terms on the LHS are the arguments of (1+3i), (1+5i) and (7+4i) respectively. Taking the products of these gives -130, which has an argument of pi.
Using arg(xyz) = arg(x) + arg(y) + arg(z), the result is proven.

 

[Masaken]

The terms on the LHS are the arguments of (1+3i), (1+5i) and (7+4i) respectively. Taking the products of these gives -130, which has an argument of pi.
Using arg(xyz) = arg(x) + arg(y) + arg(z), the result is proven.

oh i didn't know you had to find the complex numbers which makes so much sense haha, thanks so much