Let's try:
if
tanα=QP, then
Qcos(2α)+Pcos(2α)=cos(2α)(Q+P)=Q.
First, try to check whether it is true substituting real numbers. Take α=30∘ for example:
tan30∘=31, or P=1, Q=3. Now substitute these values into what we must prove:
cos(2⋅30∘)(3+1)=1.366, and it is not equal to Q=3.
Perhaps there is a typo and instead of what we wrote we must write
Qcos2(α)+Pcos2(α)=cos2(α)(Q+P)=Q. Check this with the same angle:
cos2(30∘)(3+1)=2.049, which again has nothing common with Q=3=1.732.
You can try it with other combination of cosines squared or with double angles and the closest result can be obtained when
Qcos2(α)+Pcos(2α)=Q:
3cos2(30∘)+Pcos(2⋅30∘)=1.799, but it is again different from Q.
This cannot be proved.
Comments