It's just 2 cut off with 2 right?

Geometry Level 2

Mandy was learning trigonometric identities one day and erroneously decided that 2 sin ⁡ ( θ 2 ) = sin ⁡ θ 2\sin\left(\frac\theta2\right)=\sin\theta was true for all θ \theta . What is the sum of all values of θ \theta in the interval [ 0 ∘ , 36 0 ∘ ] [0^\circ,360^\circ] for which this equation is actually true?


The answer is 360.

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Elliott Macneil by Elliott Macneil , 22, United Kingdom

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3 solutions

Simon Esslemont
Jun 21, 2015

2sin(0.5x) has twice the period of sin(x). So they start together at 0 and finish at together 360. Hence sum is 0+360=360.

2sin(0.5x) would be zero at 0 and 360. But the 2"co-efficient" causes the plot to stay well above any value of sin(x), that is until it comes down to zero at 360.

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Edwin Hughes
Jun 21, 2015

2 sin ⁡ x 2 = sin ⁡ x 2\sin{\frac{x}{2} }=\sin{x}

Using the principal root from the half-angle sine identity yields:

2 1 − cos ⁡ x 2 = sin ⁡ x 2\sqrt{ \frac{1-\cos{x}}{2} }=\sin{x}

Square both sides (be weary of extraneous roots!):

4 ( 1 − cos ⁡ x 2 ) = sin ⁡ 2 x 4 \big( \frac{1-\cos{x}}{2} \big) =\sin^2{x}

2 ( 1 − cos ⁡ x ) = 1 − cos ⁡ 2 x 2 \big( 1-\cos{x}\big) =1-\cos^2{x}

2 − 2 cos ⁡ x = 1 − cos ⁡ 2 x 2-2\cos{x} =1-\cos^2{x}

cos ⁡ 2 x − 2 cos ⁡ x + 1 = 0 \cos^2{x}-2\cos{x}+1 =0

( cos ⁡ x − 1 ) 2 = 0 \big(\cos{x}-1\big)^2=0

cos ⁡ x = 1 \cos{x}=1

∴ x = 0 ∘ \therefore x=0^{\circ} and x = 36 0 ∘ x=360^{\circ}

Checking both answers in the original equation verifies that there are no extraneous roots.

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That's almost exactly how I did it! :) only on line 3 (or 4 depending how you look at it) you could use the difference of two squares to make it neater :) great solution nonetheless!

Elliott Macneil - 5 years, 11 months ago

You've forgot that

sin ⁡ x 2 = ± 1 − cos ⁡ x 2 \sin\frac{x}{2}=\pm\sqrt\frac{1-\cos x}{2}

so

2 sin ⁡ x 2 = ± 2 1 − cos ⁡ x 2 2\sin\frac{x}{2}=\pm 2 \sqrt\frac{1-\cos x}{2}

and this is the reason because you've not any extraneous solutions...

Ernesto Civello - 5 years, 11 months ago

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The negative expression produces exactly the same scenario -- after squaring -- that I outlined above (which is why I omitted it and used the principal square root). The "omission" of the negative expression has nothing to do with encountering extraneous roots.

Edwin Hughes - 5 years, 11 months ago

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I know it... and this is the reason because the squaring doesn't produce any weird solution!

Ernesto Civello - 5 years, 11 months ago
Elliott Macneil
Jun 15, 2015

Sorry everyone, this is my first problem published, and I know absolutely no LaTeX whatsoever, if someone could try to teach me, or reformat the problem, I'd be very grateful.

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It has been converted.

Don't need to apologize, everyone is a beginner at some point.

Here's a good beginner's guide to learn LaTeX \LaTeX : Beginner LaTeX Guide .

Can you provide a proper solution to your problem?

Brilliant Mathematics Staff - 5 years, 12 months ago

No problem. I think that the sum is 0 but my answer is checked as wrong. Why?

Ernesto Civello - 5 years, 12 months ago

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You forgot 360

Simon Esslemont - 5 years, 11 months ago

You basically need to reduce the problem to cos ⁡ θ = 1 \cos\theta\ = 1 and sum the values after that. 0 works, but so does 360, and hence when you sum the values you get the above answer.

Elliott Macneil - 5 years, 12 months ago

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Excuse me, but I need to reduce to

cos ⁡ ( θ 2 ) = 1 \cos (\frac{\theta}{2}) =1 because sin ⁡ ( θ ) = sin ⁡ ( 2 ⋅ θ 2 ) = 2 ⋅ sin ⁡ θ 2 cos ⁡ θ 2 \sin (\theta) = \sin (2\cdot\frac{\theta}{2}) = 2\cdot\sin\frac{\theta}{2}\cos\frac{\theta}{2}

2 ⋅ sin ⁡ θ 2 cos ⁡ θ 2 = 2 ⋅ sin ⁡ ( θ 2 ) 2\cdot\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cdot\sin(\frac{\theta}{2}) only when cos ⁡ ( θ 2 ) = 1 \cos (\frac{\theta}{2}) =1 or when 2 ⋅ sin ⁡ θ 2 = 0 2\cdot\sin\frac{\theta}{2}=0

The first equation has one solution in θ = 0 ° \theta=0° while the second one have two solutions in θ 2 = 0 ⇒ θ = 0 \frac{\theta}{2}=0\Rightarrow\theta=0 and θ 2 = 180 ° ⇒ θ = 360 ° \frac{\theta}{2}=180°\Rightarrow\theta=360°

So the solution 360° is right, but your hint... is wrong ;-)

Ernesto Civello - 5 years, 12 months ago

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@Ernesto Civello – But if θ = 360 \theta\ = 360 , then surely cos ⁡ θ 2 \cos\frac{\theta}{2} gives -1, not 1. Maybe I missed something in your note, but I think you're wrong.

Elliott Macneil - 5 years, 12 months ago

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@Elliott Macneil – Maybe I am just missing something....

Elliott Macneil - 5 years, 12 months ago

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@Elliott Macneil – Ok. I've found my error so I try to explain:

2 sin ⁡ ( θ 2 ) = sin ⁡ ( θ ) 2\sin(\frac{\theta}{2})=\sin(\theta)

2 sin ⁡ ( θ 2 ) = sin ⁡ ( 2 ⋅ θ 2 ) 2\sin(\frac{\theta}{2})=\sin(2\cdot\frac{\theta}{2})

2 sin ⁡ ( θ 2 ) = 2 sin ⁡ ( θ 2 ) cos ⁡ ( θ 2 ) 2\sin(\frac{\theta}{2})=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})

a . sin ⁡ ( θ 2 ) ≠ 0 → θ 2 ≠ 0 ° , 180 ° → θ ≠ 0 ° , 360 ° a. {\sin(\frac{\theta}{2}) \neq 0} \rightarrow \frac{\theta}{2}\neq 0°, 180° \rightarrow \theta\neq 0°, 360°

(divide by sin ⁡ ( θ 2 ) \sin(\frac{\theta}{2}) )

cos ⁡ ( θ 2 ) = 1 \cos(\frac{\theta}{2})=1

a . 1 θ 2 = 0 ° → θ = 0 ° a.1\frac{\theta}{2}=0° \rightarrow \theta = 0° (in contrast with the hypothesis)

a . 2 θ 2 = 360 ° → θ = 720 ° a.2\frac{\theta}{2}=360° \rightarrow \theta = 720° (out of range)

b . sin ⁡ ( θ 2 ) = 0 b. {\sin(\frac{\theta}{2}) = 0}

θ 2 = 0 ° , 180 ° → θ = 0 ° , 360 ° ( O K ) \frac{\theta}{2}= 0°, 180° \rightarrow \theta= 0°, 360° (OK)

0 ° + 360 ° = 360 ° 0°+360° = 360°

Ernesto Civello - 5 years, 12 months ago

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@Ernesto Civello – Cool, I think my solution might be different, but I totally missed your solution, it's certainly quite nice! :)

Elliott Macneil - 5 years, 12 months ago

This will help you learn LaTeX \LaTeX . http://www.codecogs.com/latex/eqneditor.php
Also, on Brilliant, when you put in the LaTeX \LaTeX commands, you need to encase them in \ ( and \ ) (without spaces in between the \ and )

Hobart Pao - 5 years, 11 months ago

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Thanks for the advice everyone! I'll be sure to check that site out @hobart pao, and be sure to check out my new problems, when they come out! :)

Elliott Macneil - 5 years, 11 months ago

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