# I Is this a Trigonometric Identity?

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1. Jul 7, 2016

### ecastro

I have encountered this equation:

$\cos^2 \gamma = \cos^2 \alpha \cdot \cos^2 \beta$

According to the paper, this is a trigonometric identity, but this is the first time I have encountered this. The angles $\alpha$ and $\beta$ are somewhat similar to the components of the distance formula, and $\gamma$ is the resultant distance.

2. Jul 7, 2016

### micromass

There's nothing we can do to help you if you don't tell us what $\alpha$, $\beta$ and $\gamma$ are.

3. Jul 7, 2016

### ecastro

I apologize. Here is a figure to help you what the angles are:

The angles $\alpha$ and $\beta$ are the angles $X$ and $Y$, respectively, in this figure. Whereas,

the angle $\gamma$ is the angle $\theta$ for this figure.

4. Jul 7, 2016

### Staff: Mentor

SM is orthogonal to MO and MP?

Set the distance (SM) to 1 - this simplifies equations but doesn't change the general case.

I don't understand the second sketch, let A be the point where the side between x and y hits the plane shown in the first sketch. Then (MA) = tan(x) and (AP) = tan(y)/cos(x) and $\sqrt{(MA)^2 + (AP)^2}$ = (MP) = tan($\gamma$). See if that simplifies to the equation you have.

5. Jul 7, 2016

### Staff: Mentor

If I understood you right then all three angles are those at the top of the pyramide. My calculation with $\overline{MS} = H = 1$ and both sides $= \frac{1}{2}$ of the rectangle in the plane with $M$ and $P$ gave me:
$$\cos α = \frac{2}{\sqrt{5}} \;,\; \cos β = \frac{9}{10} \;,\; \cos γ = \frac{3}{4}$$
which fails the given equation.

6. Jul 7, 2016

### Staff: Mentor

Something went wrong in your calculation then, because the equation is right - at least with the assumptions I made in my post.

I get a different result for $\cos(\beta)$ and $\cos(\gamma)$.

7. Jul 7, 2016

### Staff: Mentor

Yes, I found my mistake: wrong assumption on the triangle with β. But the general case still gives me some headache.

8. Jul 7, 2016

### Staff: Mentor

I calculated the approach I suggested, it leads to the identity in post 1 within a few lines.

9. Jul 7, 2016

### Staff: Mentor

Got it. I didn't understand your equation for $(AP)$ but the law of cosines with some Pythagoras did the job.

10. Jul 7, 2016

### Staff: Mentor

1/cos(x) is the length of the side where the angle beta is. No magic involved.

11. Jul 7, 2016

### Staff: Mentor

Sure. I meant the tan(y)-part. But I didn't really try, since the cosine law did it. After my first false assumption on height and bisection of this triangle, I picked the obvious formula.

12. Jul 7, 2016

### robphy

13. Jul 8, 2016

### ecastro

I tried simplifying this equation:

\begin{eqnarray*}
\overline{MA}^2 + \overline{AP}^2 &=& \overline{MP}^2 \\
\tan^2 x + \frac{\tan^2 y}{\cos^2 x} &=& \tan^2 \theta \\
\cos^2 x \tan^2 x + \tan^2 y &=& \cos^2 x \tan^2 \theta \\
\sin^2 x + \tan^2 y &=& \cos^2 x \tan^2 \theta
\end{eqnarray*}

I am already stuck here... I'm not really good at this, I'm sorry.

14. Jul 8, 2016

### Staff: Mentor

Use $\sin^2(x) = 1-\cos^2(x)$ to get rid of all sines (also those hidden in the tangents). To simplify the expressions, it is useful to introduce new variables like $\cos^2(x)=a$, $\cos^2(y)=b$, $\cos^2(\theta)=c$.

15. Jul 8, 2016

### Staff: Mentor

I still don't have the $\overline{AP}$ part, but I can show you my approach.

$\cos x = \frac{\overline{SM}}{\overline{AS}} \; , \; \cos θ = \frac{\overline{SM}}{\overline{SP}}$ so we have to show that $\cos y = \frac{\overline{AS}}{\overline{SP}}$.

The law of cosine here is $$\cos y = \frac{\overline{AP}^2-\overline{AS}^2-\overline{SP}^2}{-2 \cdot \overline{AS} \cdot \overline{SP}}.$$
With the Pythagoras $\overline{AP}^2 = \overline{MP}^2 - \overline{AM}^2$ and $\overline{MP}^2 = \overline{SP}^2-\overline{SM}^2$ and $\overline{AM}^2 = \overline{AS}^2 - \overline{SM}^2$ the result follows.

Modulo me haven't made any typos.

16. Jul 8, 2016

### Staff: Mentor

Wait, isn't that circular? You use the law of cosines to prove that cos y is indeed the ratio of two angles in the triangle.

Actually, the three cosine definitions are sufficient to show the identity. The whole problem can be solved in two lines.

17. Jul 8, 2016

### Staff: Mentor

I don't think so. Basically I show that the law of cosine is equivalent to the statement that has to be proven. I simply convert all angles to (relations of) straight lines from $S.$ Pythagoras only substitutes variables and for the cosines of $x$ and $θ$ I use their definitions.
I believe you but I can't see it. Probably it will be embarrassing as soon as I understood your solution. I am already stuck at $\overline{AP} = \frac{\tan y}{\cos x}.$

Edit: But it is right that it is somehow suspicious that I don't need the squares.

Last edited: Jul 8, 2016
18. Jul 8, 2016

### Staff: Mentor

The derivation works for any M, and does not include the other angles at all. It is just a very complicated way to simplify the law of cosines for a rectangular triangle involving additional irrelevant steps.

Draw the triangle SMA, with a right angle at M. Then (SM) = (SA) cos x.
Draw the triangle SAP, with a right angle at A. Then (SA) = (SP) cos y.
Draw the triangle SMP, with a right angle at M. Then (SM) = (SP) cos $\theta$.

This is assuming x,y,$\theta < \frac{\pi}{2}$, otherwise you cannot have the right triangles. Larger angles just lead to sign flips, squaring the cosines gets rid of those details.

19. Jul 8, 2016

### Staff: Mentor

Thanks, the right angle at $A$ in $SAP$ was what I was missing. I told you it would be embarrassing.

20. Jul 8, 2016

### Staff: Mentor

The equation for the cosine would be wrong otherwise - it works exactly if the angle at A in SAP is a right one.

The right angle can be derived from the two right angles at M (in SMA and SMP) and the right angle at A in MAP, so it is nothing you have to plug in manually.