Let's try:
if
tan α = P Q , \text{tan}\alpha=\frac{P}{Q}, tan α = Q P , then
Q cos ( 2 α ) + P cos ( 2 α ) = cos ( 2 α ) ( Q + P ) = Q . Q\text{cos}(2\alpha)+P \text{cos}(2\alpha)=\text{cos}(2\alpha)(Q+P)=Q. Q cos ( 2 α ) + P cos ( 2 α ) = cos ( 2 α ) ( Q + P ) = Q .
First, try to check whether it is true substituting real numbers. Take α = 3 0 ∘ \alpha=30^\circ α = 3 0 ∘ for example:
tan 3 0 ∘ = 1 3 , \text{tan}30^\circ=\frac{1}{\sqrt{3}}, tan 3 0 ∘ = 3 1 , or P = 1 , Q = 3 . P=1,\space\space\space Q=\sqrt{3}. P = 1 , Q = 3 . Now substitute these values into what we must prove:
cos ( 2 ⋅ 3 0 ∘ ) ( 3 + 1 ) = 1.366 , \text{cos}(2\cdot30^\circ)(\sqrt{3}+1)=1.366, cos ( 2 ⋅ 3 0 ∘ ) ( 3 + 1 ) = 1.366 , and it is not equal to Q = 3 . Q=\sqrt{3}. Q = 3 .
Perhaps there is a typo and instead of what we wrote we must write
Q cos 2 ( α ) + P cos 2 ( α ) = cos 2 ( α ) ( Q + P ) = Q . Q\text{cos}^2(\alpha)+P \text{cos}^2(\alpha)=\text{cos}^2(\alpha)(Q+P)=Q. Q cos 2 ( α ) + P cos 2 ( α ) = cos 2 ( α ) ( Q + P ) = Q . Check this with the same angle:
cos 2 ( 3 0 ∘ ) ( 3 + 1 ) = 2.049 , \text{cos}^2(30^\circ)(\sqrt{3}+1)=2.049, cos 2 ( 3 0 ∘ ) ( 3 + 1 ) = 2.049 , which again has nothing common with Q = 3 = 1.732. Q=\sqrt{3}=1.732. 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
Q cos 2 ( α ) + P cos ( 2 α ) = Q : Q\text{cos}^2(\alpha)+P \text{cos}(2\alpha)=Q: Q cos 2 ( α ) + P cos ( 2 α ) = Q :
3 cos 2 ( 3 0 ∘ ) + P cos ( 2 ⋅ 3 0 ∘ ) = 1.799 , \sqrt{3}\text{cos}^2(30^\circ)+P \text{cos}(2\cdot30^\circ)=1.799, 3 cos 2 ( 3 0 ∘ ) + P cos ( 2 ⋅ 3 0 ∘ ) = 1.799 , but it is again different from Q.
This cannot be proved.
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