Solving trigonometry equation involving half-angle

[songoku]

Homework Statement

Please see below

Relevant Equations

Double angle formula

I can solve this by using the double-angle formula but the teacher expects another method not involving the double-angle formula.

Is there a way to solve this without using double-angle formula?

Thanks

 

[fresh_42]

Can you use the complex exponential function?

 

[kuruman]

You can start with the trig identity
$\sin(a+b)=\sin a \cos b+\cos a \sin b$
and set $a=b=\dfrac{\theta}{2}.$

 

[bob012345]

You can also iterate numerically.

 

[Orodruin]

You can start with the trig identity
$\sin(a+b)=\sin a \cos b+\cos a \sin b$
and set $a=b=\dfrac{\theta}{2}.$

That’s just giving the double angle formula though.

Without it I would go for the complex exponential solution mentioned by fresh.

 

[kuruman]

That’s just giving the double angle formula though.

Without it I would go for the complex exponential solution mentioned by fresh.

I thought about that, but the quoted trig identity can be derived directly using complex exponentials without invoking the double angle formula. Regardless of how one goes about solving this, one will have to use the given fact that the argument of the trig function on the left is half the argument of the trig function on the right.

 

[Orodruin]

I thought about that, but the quoted trig identity can be derived directly using complex exponentials without invoking the double angle formula.

Sure, but it is the double angle formula once you put both arguments to theta/2.

In general however, I do find it silly to prohibit certain paths to solve a problem unless there is very good reason to do so.

 

[kuruman]

Sure, but it is the double angle formula once you put both arguments to theta/2.

In general however, I do find it silly to prohibit certain paths to solve a problem unless there is very good reason to do so.

I agree. Prohibiting paths raises the question whether what is not explicitly prohibited is implicitly allowed. If using WolframAlpha is not prohibited, for example, then it can be used.

 

[bob012345]

What class is this in and what level of school is it?

 

[neilparker62]

Homework Statement:: Please see below
Relevant Equations:: Double angle formula

View attachment 314005

I can solve this by using the double-angle formula but the teacher expects another method not involving the double-angle formula.

Is there a way to solve this without using double-angle formula?

Thanks

$$\frac{7}{\sqrt{1+t^2}}=10\times \frac{2t}{1+t^2}$$

 

[kuruman]

$$\frac{7}{\sqrt{1+t^2}}=10\times \frac{2t}{1+t^2}$$

Yes, but how do you derive this expression for the tangent $t$ of the half-angle without using the half-angle formula on the right-hand side?

 

[neilparker62]

Yes, but how do you derive this expression for the tangent $t$ of the half-angle without using the half-angle formula on the right-hand side?

I thought that not using the double angle meant not solving: $$7\cos\left(\frac{\theta}{2}\right)=20\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$$ which would be straight forward.

 

[kuruman]

See reply #7 by @Orodruin. I agree with him that this is silly and I have said all I had to say here.

 

[bob012345]

Iterate on your calculator $$\theta = sin^{-1}\left(\frac{7}{10} cos(\frac{\theta}{2})\right)$$ until it converges.

 

[songoku]

I asked the teacher for clarification about another method he expected then he sent me a pdf email containing a lot formulas. For trigonometry part, the formula is only sine and cosine rule.

I am starting to think this is a troll rather than a serious question. I just don't see how the question is related to sine and cosine rule.

Is it just me or there is actually a way to solve this question using sine and cosine rule?

Thanks

 

[epenguin]

You could try the special angles you know sines and cosines of without wasting time on ones that obviously won't work like x=0.

 

[songoku]

Thank you very much for the help fresh_42, kuruman, bob012345, Orodruin, neilarpker62, epenguin

 

[SammyS]

$$\frac{7}{\sqrt{1+t^2}}=10\times \frac{2t}{1+t^2}$$

This looks to me like a reasonable path toward a solution.

It probably should be pointed out that in the right hand side, the tangent half-angle formula is used for $\sin \theta$, where $t=\tan (\theta / 2)$ . See https://en.wikipedia.org/wiki/Tangent_half-angle_formula

In the left hand side, $\cos(\theta / 2)$ is expressed in terms of $\tan (\theta / 2)$.

 

[neilparker62]

I asked the teacher for clarification about another method he expected then he sent me a pdf email containing a lot formulas. For trigonometry part, the formula is only sine and cosine rule.

I am starting to think this is a troll rather than a serious question. I just don't see how the question is related to sine and cosine rule.

Is it just me or there is actually a way to solve this question using sine and cosine rule?

Thanks

 

[SammyS]

@neilparker62 ,

Looks good to me, Clever !

 

[songoku]

View attachment 314156

Wow, there is really a way to solve the question using the sine and cosine rule.

Thank you very much for the help neilparker62

 

[neilparker62]

Wow, there is really a way to solve the question using the sine and cosine rule.

Thank you very much for the help neilparker62

You could even drop a perpendicular and solve by elementary trig! After which we have come full circle since:$$ \cos\left(\frac{\theta}{2}\right)\left(20\sin\left(\frac{\theta}{2}\right)-7\right)=0.$$

 

[neilparker62]

@neilparker62 ,

Looks good to me, Clever !

Thanks. Faith cometh not by sines but by cosines

 

[epenguin]

For benefit of the weaker brethren, can songaku or someone please spell out the arguments – both methods? I am not seeing it.

There are three solutions 0 ≤ θ ≤ 2π . One of them you can obtain as indicated in #16 (and you may kick yourself if you have missed it).

Actually the periodicity of the problem is 4π not 2π (Another reason for doubt as to whether the Prof who set it knew exactly what he's about) and there are two of these 'trivial' solutions, and still only two others 0 ≤ θ ≤ 4π , I think.

 

[songoku]

For benefit of the weaker brethren, can songaku or someone please spell out the arguments – both methods? I am not seeing it.

There are three solutions 0 ≤ θ ≤ 2π . One of them you can obtain as indicated in #16 (and you may kick yourself if you have missed it).

By both methods, do you mean the methods in post #19 and post #22?

If yes, for the method in post #19:
$$7 \cos \left(\frac{\theta}{2}\right)=10 \sin \theta$$
$$\frac{7}{\sin \theta}=\frac{10}{\cos \frac{\theta}{2}}$$
$$\frac{7}{\sin \theta}=\frac{10}{\sin \left(\frac{\pi}{2}-\frac{\theta}{2}\right)}$$

This is in the form of sine rule and then I changed it into triangle diagram to get the same picture as post#19

Solving the cosine rule, I got 0.72 and 5.57 as the solution. I got the third solution by using your method in post #16For the method in post #22, I used the double angle formula (which is maybe prohibited by my teacher):
$$7 \cos \left(\frac{\theta}{2}\right)=10 \sin \theta$$
$$7 \cos \left(\frac{\theta}{2}\right) = 10 (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})$$
$$7 \cos \left(\frac{\theta}{2}\right) - 10 (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})=0$$
$$ \cos\left(\frac{\theta}{2}\right)\left(7-20\sin\left(\frac{\theta}{2}\right)\right)=0$$

Solving this, I got all the three solutions

 

[bob012345]

You could even drop a perpendicular and solve by elementary trig! After which we have come full circle since:$$ \cos\left(\frac{\theta}{2}\right)\left(20\sin\left(\frac{\theta}{2}\right)-7\right)=0.$$

How do you get this from the diagram which has only one angle solution?

 

[kuruman]

To @songoku:
To elaborate on @bob012345's statement, if you want to use the triangle, you need to provide a geometrical proof. Hint: You need to find two line segments, one of length $7 \cos \left(\frac{\theta}{2}\right)$ and one of length $10 \sin \theta$ and show that they are equal.

 

[neilparker62]

How do you get this from the diagram which has only one angle solution?

How do you get this from the diagram which has only one angle solution?

The solution to the second bracket is that which you can get either by dropping a perpendicular and using elementary trig or by applying sine rule to the triangle I drew. In the latter case, I would say the solution $\cos(\frac{\theta}{2})=0$ is excluded since $\cos(\frac{\theta}{2})$ would be a denominator in the equation $\frac{7}{\sin\theta}=\frac{10}{\cos(\frac{\theta}{2})}$. But we can of course solve for $\theta$ directly in the triangle by applying the cosine rule.

 

[songoku]

To @songoku:
To elaborate on @bob012345's statement, if you want to use the triangle, you need to provide a geometrical proof. Hint: You need to find two line segments, one of length $7 \cos \left(\frac{\theta}{2}\right)$ and one of length $10 \sin \theta$ and show that they are equal.

Sorry I don't really understand your hint.

What I did was only change sine rule into triangle:
$$\frac{7}{\sin \theta}=\frac{10}{\sin \left(\frac{\pi}{2}-\frac{\theta}{2}\right)}$$

Side of length 7 will be across the angle $\theta$ and side of length 10 will be across angle $\frac{\pi}{2} - \frac{\theta}{2}$. One more angle of the triangle will be $\frac{\pi}{2} - \frac{\theta}{2}$ so the triangle is an isosceles triangle and the third side will also be 10.

In the latter case, I would say the solution $\cos(\frac{\theta}{2})=0$ is excluded since $\cos(\frac{\theta}{2})$ would be a denominator in the equation $\frac{7}{\sin\theta}=\frac{10}{\cos(\frac{\theta}{2})}$.

But the solution of $\cos(\frac{\theta}{2})=0$ satisfies the original question so wouldn't we lose one solution if we exclude that?

Thanks

 

[kuruman]

Sorry I don't really understand your hint.

Here is what I had in mind. The diagram is drawn to scale specifically for this problem but the proof is general because the proposition that we are asked to prove is an identity.

1. Draw isosceles triangle ABC.
2. Draw heights AF and BD.
3. From right triangle ADB we have $BD = AB \sin(2\alpha).$
4. From right triangle AFB BDC we have $BD = BC \cos(\beta).$
5. It follows that $AB \sin(2\alpha)=BC \cos(\beta).$
6. Angles $\alpha$ and $\beta$ are equal because their sides are mutually perpendicular. Therefore, $BC \cos(\alpha)=AB \sin(2\alpha).$
7. Setting $\alpha = \frac{\theta}{2}$, $AB=10$ and $BC=7$ finishes the proof for this specific case.

 

[kuruman]

It looks like this is for triangle BDC not AFB .

Good catch! I made the necessary correction. Thanks.

 

[neilparker62]

But the solution of $\cos(\frac{\theta}{2})=0$ satisfies the original question so wouldn't we lose one solution if we exclude that?

Thanks

Perhaps we should rather say it's a trivial solution since 7 x 0 = 10 x 0.

 

[SammyS]

Perhaps we should rather say it's a trivial solution since 7 x 0 = 10 x 0.

Well, it's a trivial solution if you use the double angle formula for sine in the equation given in the OP.

It's also apparent that it's a solution if you note that $\cos\left(\frac \theta 2 \right)=0$ for $\theta$ being an odd multiple of $\pi$ and that $\sin\left ( \theta \right )=0$ for any integer multiple of $\pi$ .

 

[songoku]

Here is what I had in mind. The diagram is drawn to scale specifically for this problem but the proof is general because the proposition that we are asked to prove is an identity.
View attachment 314276

1. Draw isosceles triangle ABC.
2. Draw heights AF and BD.
3. From right triangle ADB we have $BD = AB \sin(2\alpha).$
4. From right triangle AFB BDC we have $BD = BC \cos(\beta).$
5. It follows that $AB \sin(2\alpha)=BC \cos(\beta).$
6. Angles $\alpha$ and $\beta$ are equal because their sides are mutually perpendicular. Therefore, $BC \cos(\alpha)=AB \sin(2\alpha).$
7. Setting $\alpha = \frac{\theta}{2}$, $AB=10$ and $BC=7$ finishes the proof for this specific case.

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

Thanks

 

[kuruman]

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

Thanks

Angle $\alpha$ (right) is formed by segments AF and AC.
Angle $\beta$ is formed by segments BD and BC.

BD is perpendicular to AC and BC is perpendicular to AF. That's what "mutually perpendicular" means. You can see that the sides will be parallel to each other when you rotate AF and AC counterclockwise by 90° which makes the angles equal.

 

[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?