Question #88930
If tana=P/Q,prove that Q cos2a+P cos2a=Q
1
Expert's answer
2019-05-02T05:24:19-0400

Let's try:

if

tanα=PQ,\text{tan}\alpha=\frac{P}{Q},

then

Qcos(2α)+Pcos(2α)=cos(2α)(Q+P)=Q.Q\text{cos}(2\alpha)+P \text{cos}(2\alpha)=\text{cos}(2\alpha)(Q+P)=Q.


First, try to check whether it is true substituting real numbers. Take α=30\alpha=30^\circ for example:


tan30=13,\text{tan}30^\circ=\frac{1}{\sqrt{3}},

or P=1,   Q=3.P=1,\space\space\space Q=\sqrt{3}. Now substitute these values into what we must prove:


cos(230)(3+1)=1.366,\text{cos}(2\cdot30^\circ)(\sqrt{3}+1)=1.366,

and it is not equal to Q=3.Q=\sqrt{3}.

Perhaps there is a typo and instead of what we wrote we must write


Qcos2(α)+Pcos2(α)=cos2(α)(Q+P)=Q.Q\text{cos}^2(\alpha)+P \text{cos}^2(\alpha)=\text{cos}^2(\alpha)(Q+P)=Q.

Check this with the same angle:


cos2(30)(3+1)=2.049,\text{cos}^2(30^\circ)(\sqrt{3}+1)=2.049,

which again has nothing common with Q=3=1.732.Q=\sqrt{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:Q\text{cos}^2(\alpha)+P \text{cos}(2\alpha)=Q:

3cos2(30)+Pcos(230)=1.799,\sqrt{3}\text{cos}^2(30^\circ)+P \text{cos}(2\cdot30^\circ)=1.799,

but it is again different from Q.

This cannot be proved.


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