Answer to Question #88930 in Trigonometry for Manu Chemjong

Question

If tana=P/Q,prove that Q cos2a+P cos2a=Q

Answer 1

Let's try:

if

$$\text{tan}\alpha=\frac{P}{Q},$$

then

$$Q\text{cos}(2\alpha)+P \text{cos}(2\alpha)=\text{cos}(2\alpha)(Q+P)=Q.$$

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

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

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

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

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

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

$$Q\text{cos}^2(\alpha)+P \text{cos}^2(\alpha)=\text{cos}^2(\alpha)(Q+P)=Q.$$

Check this with the same angle:

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

which again has nothing common with $Q=\sqrt{3}=1.732.$

You can try it with other combination of cosines squared or with double angles and the closest result can be obtained when

$$Q\text{cos}^2(\alpha)+P \text{cos}(2\alpha)=Q:$$

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

but it is again different from Q.

This cannot be proved.