This question is breaking my head (2 Viewers)

upishcat · Dec 14, 2019

where do I even start???

Drdusk · Dec 14, 2019

upishcat said:
where do I even start???

It's kind of shitty how they put literally 3 questions into one part lol. Anyway

$\text{We have to show}\hspace{2mm}\dfrac{e^{i\alpha} + e^{i\beta}}{e^{i(\alpha + \beta)/2}} =2 \cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)$

$LHS = \dfrac{e^{i\alpha} + e^{i\beta}}{e^{i(\alpha + \beta)/2}} = \dfrac{\cos\alpha + \cos\beta + i(\sin\alpha + \sin\beta)}{e^{i(\alpha + \beta)/2}}$

$\text{Now remember the formulae for cos and sin}\hspace{2mm}\cos\alpha + \cos\beta = 2 \cos \bigg(\dfrac{\alpha + \beta}{2}\bigg) \cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)$

$\therefore LHS = \dfrac{2 \cos \bigg(\dfrac{\alpha + \beta}{2}\bigg) \cos \bigg(\dfrac{\alpha - \beta}{2}\bigg) + i\Bigg(2 \sin \bigg(\dfrac{\alpha + \beta}{2}\bigg) \cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\Bigg)}{e^{i(\alpha + \beta)/2}}$

$= 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\times \dfrac{\cos \bigg(\dfrac{\alpha + \beta}{2}\bigg) + i\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)}{e^{i(\alpha + \beta)/2}}$

$= 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\times\dfrac{e^{i(\alpha + \beta)/2}}{e^{i(\alpha + \beta)/2}}$

$\text{Which thus gives the required result.}$

upishcat · Dec 14, 2019

Im literally woke after seeing your explanation. Thank you so much my dude.

blizzardblizzard · Dec 14, 2019

woah is euler's formula in the new 4u syllabus now?

upishcat · Dec 14, 2019

blizzardblizzard said:
woah is euler's formula in the new 4u syllabus now?

Yep! I personally think it a great addition. It truly is fascinating how Euler combined so many fundamental numbers in math and made a forumula out of them.

Drdusk · Dec 14, 2019

upishcat said:
Im literally woke after seeing your explanation. Thank you so much my dude.

I just got back home so here's the rest.

$\text{Now we know}\hspace{2mm}e^{i\alpha} + e^{i\beta} = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)e^{i(\alpha + \beta)/2}$

$\therefore \cos\alpha + \cos\beta+ i( \sin\alpha + \sin\beta) = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)e^{i(\alpha + \beta)/2}$

$\text{Now observe that we need to find the imaginary part of the RHS so we can equate it to the imaginary part of the LHS}$

$\therefore RHS = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\Bigg(\cos(\alpha + \beta)/2+i\sin(\alpha + \beta)/2\Bigg)$

$\text{Now we take the imaginary part of the RHS as we can see we need sin(alpha) + sin(beta) in the numerator}$

$\text{Im}(RHS) =2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)$

$\therefore \sin\alpha + \sin\beta = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)$

$\therefore \dfrac{\sin\alpha + \sin\beta}{\cos\alpha + \cos\beta} = \dfrac{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)}{\cos\alpha + \cos\beta}$

$=\dfrac{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)}{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\cos \bigg(\dfrac{\alpha + \beta}{2}\bigg)}$

$= \tan\bigg(\dfrac{\alpha + \beta}{2}\bigg)$

Drdusk · Dec 14, 2019

upishcat · Dec 14, 2019

Drdusk said:
I just got back home so here's the rest.

$\text{Now we know}\hspace{2mm}e^{i\alpha} + e^{i\beta} = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)e^{i(\alpha + \beta)/2}$

$\therefore \cos\alpha + \cos\beta+ i( \sin\alpha + \sin\beta) = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)e^{i(\alpha + \beta)/2}$

$\text{Now observe that we need to find the imaginary part of the RHS so we can equate it to the imaginary part of the LHS}$

$\therefore RHS = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\Bigg(\cos(\alpha + \beta)/2+i\sin(\alpha + \beta)/2\Bigg)$

$\text{Now we take the imaginary part of the RHS as we can see we need sin(alpha) + sin(beta) in the numerator}$

$\text{Im}(RHS) =2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)$

$\therefore \sin\alpha + \sin\beta = 2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)$

$\therefore \dfrac{\sin\alpha + \sin\beta}{\cos\alpha + \cos\beta} = \dfrac{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)}{\cos\alpha + \cos\beta}$

$=\dfrac{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\sin \bigg(\dfrac{\alpha + \beta}{2}\bigg)}{2\cos \bigg(\dfrac{\alpha - \beta}{2}\bigg)\cos \bigg(\dfrac{\alpha + \beta}{2}\bigg)}$

$= \tan\bigg(\dfrac{\alpha + \beta}{2}\bigg)$

But when we take the Im(RHS) shouldnt we not consider 2cos(a-b/2) because that is a real part?

Drdusk · Dec 14, 2019

upishcat said:
But when we take the Im(RHS) shouldnt we not consider 2cos(a-b/2) because that is a real part?

I did consider it. The Im(RHS) line in mine has the 2cos(a-b/2) in it..

upishcat · Dec 14, 2019

Drdusk said:
$\text{If we want to find tan of 5pi/24 we can see if we set alpha and beta as 3pi/12 and 2pi/12 we get 5pi/24}$

$\therefore \alpha = \pi/4, \beta = \pi/6$

$\therefore \tan\bigg(\dfrac{5\pi}{24}\bigg) = \dfrac{2 + \sqrt{2}}{2 + \sqrt{6}}$

May i ask how you got this? I got as far as to understanding that 3pi/12 and 2pi/12 are used because we know their values using pythagoras triangles

Drdusk · Dec 14, 2019

upishcat said:
May i ask how you got this? I got as far as to understanding that 3pi/12 and 2pi/12 are used because we know their values using pythagoras triangles

It's from the formula we obtained for tan(alpha + beta \2). You just plug in the values of alpha and beta into the sin and cos equation.

upishcat · Dec 14, 2019

Drdusk said:
It's from the formula we obtained for tan(alpha + beta \2). You just plug in the values of alpha and beta into the sin and cos equation.

Ooops. How dumb of me haha.

upishcat · Dec 14, 2019

S

upishcat said:
May i ask how you got this? I got as far as to understanding that 3pi/12 and 2pi/12 are used because we know their values using pythagoras triangles

shouldn't you completely rationalise the denominator to get surd form?

HeroWise · Dec 14, 2019

You are a boomer if u like Eulers form

DoomieDaJuan · Dec 14, 2019

HeroWise said:
You are a boomer if u like Eulers form

ok boomer

HeroWise · Dec 14, 2019

I, Giorno Giovanna, hate eulers form!

Drdusk · Dec 14, 2019

HeroWise said:
I, Giorno Giovanna, hate eulers form!

Wait until uni. All the boomers will be pissed off if they even see cis. At least that's been my experience imao.

HeroWise · Dec 14, 2019

interestingly I want to be a farmer when i grow up. So no uni for me amirite.
Jokes aside, aah. Interestingly, my career path atm looks really unrelated to maths rip.

stupid_girl · Dec 14, 2019

I guess the question is not expecting you to apply trigonometric identities for product to sum or sum to product. Otherwise, there's no point to bring complex number into the question. Having said that, I think a trigonometric approach is more elegant, especially for the second part of the question.

$\frac{e^{i\alpha}+e^{i\beta}}{e^{i\frac{\alpha+\beta}{2}}}=\frac{e^{i\alpha}}{e^{i\frac{\alpha+\beta}{2}}}+\frac{e^{i\beta}}{e^{i\frac{\alpha+\beta}{2}}}=e^{i\frac{\alpha-\beta}{2}}+e^{i\frac{\beta-\alpha}{2}}=\cos\frac{\alpha-\beta}{2}+i\sin\frac{\alpha-\beta}{2}+\cos\frac{\beta-\alpha}{2}+i\sin\frac{\beta-\alpha}{2}=2\cos\frac{\alpha-\beta}{2}$

$e^{i\alpha}+e^{i\beta}=2\cos\frac{\alpha-\beta}{2}e^{i\frac{\alpha+\beta}{2}}$

$\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}=\frac{Im\left(e^{i\alpha}+e^{i\beta}\right)}{Re\left(e^{i\alpha}+e^{i\beta}\right)}=\frac{2\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}}=\tan\frac{\alpha+\beta}{2}$

upishcat · Dec 15, 2019

stupid_girl said:
I guess the question is not expecting you to apply trigonometric identities for product to sum or sum to product. Otherwise, there's no point to bring complex number into the question. Having said that, I think a trigonometric approach is more elegant, especially for the second part of the question.

$\frac{e^{i\alpha}+e^{i\beta}}{e^{i\frac{\alpha+\beta}{2}}}=\frac{e^{i\alpha}}{e^{i\frac{\alpha+\beta}{2}}}+\frac{e^{i\beta}}{e^{i\frac{\alpha+\beta}{2}}}=e^{i\frac{\alpha-\beta}{2}}+e^{i\frac{\beta-\alpha}{2}}=\cos\frac{\alpha-\beta}{2}+i\sin\frac{\alpha-\beta}{2}+\cos\frac{\beta-\alpha}{2}+i\sin\frac{\beta-\alpha}{2}=2\cos\frac{\alpha-\beta}{2}$

$e^{i\alpha}+e^{i\beta}=2\cos\frac{\alpha-\beta}{2}e^{i\frac{\alpha+\beta}{2}}$

$\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}=\frac{Im\left(e^{i\alpha}+e^{i\beta}\right)}{Re\left(e^{i\alpha}+e^{i\beta}\right)}=\frac{2\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}}=\tan\frac{\alpha+\beta}{2}$

Sorry DrDusk but I think I will have to go with this one. Stupid_girl's got a point.

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