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Working with LHS:

I've tried using the sum to product trig ID to get:

$1 + 2\cos(3x/2)\cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.

I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.

Andrei
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McMath
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4 Answers4

1

Start with the RHS. Notice that

$$ \sin(5x/2) = \sin(2x + x/2) = \sin 2x \cos (x/2) + \cos 2x \sin (x/2) $$

Also, notice that

$$ \sin 2x = 2 \sin x \cos x = 4 \sin (x/2) \cos (x/2) \cos x $$

Therefore,

$$ \frac{ \sin (5x/2) }{2 \sin(x/2) } = \frac{4 \sin (x/2) \cos^2 (x/2) \cos x + \cos 2x \sin (x/2)}{2 \sin(x/2)} $$

$$ = 2 \cos^2 (x/2) \cos x + \frac{ \cos 2x }{2} $$

Also, we have that $\cos^2 (x/2) = \frac{ \cos x + 1 }{2}$ and so

$$ = \cos^2 x + \cos x + \frac{ \cos 2x }{2} $$

$$ = \frac{1+\cos 2x }{2} + \cos x + \frac{ \cos 2x }{2} $$

$$ \cos x + \cos 2x + \frac{1}{2} $$$

add the missing $1/2$ from the RHS and you have the LHS

James
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  • How did you get that cos^2(x/2) = (cosx + 1)/2 – McMath Jan 16 '19 at 04:22
  • double angle formula $ x = 2 \cdot \frac{x}{2} $ – James Jan 16 '19 at 04:24
  • hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1? – McMath Jan 16 '19 at 04:29
1

For $\sin\frac{x}{2}\neq0$ we obtain: $$1+\cos{x}+\cos2x=\frac{2\sin\frac{x}{2}+2\sin\frac{x}{2}\cos{x}+2\sin\frac{x}{2}\cos2x}{2\sin\frac{x}{2}}=$$ $$=\frac{2\sin\frac{x}{2}+\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{1}{2}+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$ I used the following formula. $$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta)).$$ For example,$$2\sin\frac{x}{2}\cos{x}=2\cdot\frac{1}{2}\left(\sin\left(\frac{x}{2}+x\right)+\sin\left(\frac{x}{2}-x\right)\right)=\sin\frac{3x}{2}-\sin\frac{x}{2}.$$

Michael Rozenberg
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Hint:

Use http://mathworld.wolfram.com/WernerFormulas.html

$2\sin\dfrac x2\cos mx=\sin\left(m+\dfrac12\right)x-\sin\left(m-\dfrac12\right)x$

Set $m=1,2$ and add to find $$2\sin\dfrac x2(\cos x+\cos2x)=\sin\dfrac{5x}2-\sin\dfrac x2$$

Assuming $\sin\dfrac x2\ne0,$ divide both sides by $2\sin\dfrac x2$

See also: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

lab bhattacharjee
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The left-hand side is $$\Re(1+e^{ix}+e^{2ix})=\Re\frac{e^{3ix}-1}{e^{ix}-1}=\Re\frac{2ie^{3ix/2}\sin\frac{3x}{2}}{2ie^{ix/2}\sin\frac{x}{2}}=\Re\frac{2e^{ix}\sin\frac{3x}{2}}{2\sin\frac{x}{2}}\\=\frac{2\cos x\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{\sin\frac{x}{2}+\sin\frac{5x}{2}}{2\sin\frac{x}{2}}=\frac12+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$

J.G.
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