Solving trigonometry equation involving half-angle

[SammyS]

Sorry, for step no. 6, what is the meaning of "their sides are mutually perpendicular" ?

Thanks

Alternatively: Notice that triangle AFC and triangle BDC are similar, so $\alpha = \beta$ .

 

[songoku]

Thank you very much for the help and explanation kuruman, neilparker62, SammyS

 

[epenguin]

We should not deny ourselves the insights that graphs give – here both Cartisian and polar presentations.

https://www.physicsforums.com/attachments/314377

The first gives, I think, some extra clarification on the two 'zero' roots. But there is something that still eludes me. I would expect by a change of variable to e.g. (π - θ) to get a symmetrical expression with a perfect square, so that the solutions are ± the same quantity. I have not been able to do this and need to go back to school - maybe I have arrived at the right place.

 

[songoku]

The first gives, I think, some extra clarification on the two 'zero' roots. But there is something that still eludes me. I would expect by a change of variable to e.g. (π - θ) to get a symmetrical expression with a perfect square, so that the solutions are ± the same quantity.

Maybe it is not possible to get an equation involving perfect square and the ± solutions are due to the symmetry of the trigonometry function?

 

[kuruman]

It seems to me that $\cos(\frac{\theta}{2})=0$ is not a solution to a particular equation. The identity $$2\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})=\sin(\theta)$$can be derived geometrically from the triangle ABC in post #30 shown slightly modified below.

We have

1. $(BC)=2(BF)=2(AB)\sin(\alpha)\implies (BC)\cos(\alpha)=2(AB)\sin(\alpha)\cos(\alpha)$
2. $(BD)=(AB)\sin(2\alpha)=(BC)\cos(\alpha)$

From 1 and 2 it follows that $$(AB)\sin(2\alpha)=2(AB)\sin(\alpha)\cos(\alpha)\tag{1}$$which proves the identity. Being an identity means that it is not an equation to be solved and that it has no roots. It means that if you put any specific value of $\alpha$ in each side and evaluate separately, you get $LHS=RHS$. That's not how equations with roots behave.

When $\alpha$, which is the same as $\frac{\theta}{2}$, is equal to 90° we have
$LHS=(AB)\sin(2\times 90^o)=(AB)\sin(180^o)=(AB)*0=0.$
$RHS=(AB)\cos( 90^o)\sin( 90^o)=(AB)*0*1=0.$

 

[Orodruin]

It seems to me that $\cos(\frac{\theta}{2})=0$ is not a solution to a particular equation.

It is a solution to the original equation stated in the OP.

 

[SammyS]

We should not deny ourselves the insights that graphs give – here both Cartisian and polar presentations.

( Graphs )

The first gives, I think, some extra clarification on the two 'zero' roots. But there is something that still eludes me. I would expect by a change of variable to e.g. (π - θ) to get a symmetrical expression with a perfect square, so that the solutions are ± the same quantity. I have not been able to do this and need to go back to school - maybe I have arrived at the right place.

Yes, the graphs are very helpful.

I'm not sure about what you mean by getting a symmetrical expression with a perfect square. As for the change of variable to (π - θ):

Letting $\gamma = (\pi - \theta )$ gives the following.

$\displaystyle 7\sin(\dfrac \gamma 2 ) = 10 \sin(\gamma), \text{ where } -\pi\le \gamma \le \pi \ .$

This version of the equation does have some nice symmetry.

 

[epenguin]

I'm not sure about what you mean by getting a symmetrical expression with a perfect square.

It was only a vague preliminary working suspicion: since there are various trigonometrical identities involving squares I suspected the solution might involve a quadratic equation. These have two solutions. If you choose the origin appropriately you get a perfect square. I was trying to find one by a geometrical construction.

The oversight, you could call it, is that squaring is not the only function that is 1:1 but has a non-unique inverse - as pointed out by songaku #39.

I first looked for simple known standard angles and hit on almost immediately
θ = π .

I could not find any construction relating 7/20 to some standard known angle, so did the same derivation of the equation
$$cos(θ/2)(7-20sin(θ/2))=0 $$
as Neilparker #12 or songaku #25.

This equation is satisfied by θ = π as already realized. Additionally we have to find solutions that satisfy
$$7-20sin(θ/2)=0 $$

These have not been explicitly stated yet here. The solutions for θ are
θ = 2 sin^-1(7/20)
My function plotter gives me a graph like this for the function sin^-1

(and at school before we had such things my tables only went up to 90° or π/2 radians).

But we know we have to extend the range and will get two values as we see in here.

In this case if $x$ is a solution then $(π - x)$ is another solution. In the present case I get solutions approximately θ = 0.7151 and 5.568.

I think if I had solved it by just writing formulae without plotting graphs I would not have noticed the second solution. Isn't it a bit worrying that such an oversight could be a small part of a computer program or of these 300 page long proofs that mathematicians do these days?