I came up with the following example on a textbook I was given.
Let's suppose we have an R-L load. I want to calculate the active power P when the voltage and the current applied on the load are given with the following functions: $$u(t)=200 \sqrt2\sin(ωt)+ 200 \sqrt2\sin(5ωt+30^{\circ}) $$ $$i(t)=26.19\sin(ωt-68.2^{\circ})+ 6.22\sin(5ωt-85.4^{\circ}) $$
In the answer it is given that we get: $$ P = \tilde{V_1} \tilde{I_1} cosφ_1 + \tilde{V_5} \tilde{I_5} cosφ_5 \Rightarrow $$ $$ P = 200\frac{26.19}{\sqrt2}cos(68.2^{\circ}) + 200\frac{6.22}{\sqrt2}cos(85.4^{\circ}) $$
What I don't get is why the angle φ5 is not calculated as : $$ φ_5 = 30^{\circ} - (-85.4) = 115.4 ^{\circ} $$ since $$ φ_n = θ_v - θ_I $$ where θv is the angle of the voltage harmonic and θi is the angle of the current harmonic. Isn't this the right was to calculate it or am I missing something?