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.


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,$$2\sin\frac{x}{2}\cos{x}=2\cdot\frac{1}{2}\left(\sin\left(\frac{x}{2}+x\right)+\sin\left(\frac{x}{2}-x\right)\right)=\sin\frac{3x}{2}-\sin\frac{x}{2}.$$

  • $\begingroup$ Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx $\endgroup$ – McMath Jan 16 at 14:42
  • $\begingroup$ @McMath I added something. See now. $\endgroup$ – Michael Rozenberg Jan 16 at 14:47
  • $\begingroup$ oh, and then you're putting all over common denom? I'll give it a shot $\endgroup$ – McMath Jan 16 at 14:57
  • $\begingroup$ By the same way: $\cos{x}=\frac{2\sin\frac{x}{2}\cos{x}}{2\sin\frac{x}{2}}$. With $\cos2x$ we make the similar thing. $\endgroup$ – Michael Rozenberg Jan 16 at 14:58
  • $\begingroup$ Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place? $\endgroup$ – McMath Jan 16 at 15:06

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

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


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

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

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

  • $\begingroup$ How did you get that cos^2(x/2) = (cosx + 1)/2 $\endgroup$ – McMath Jan 16 at 4:22
  • $\begingroup$ double angle formula $ x = 2 \cdot \frac{x}{2} $ $\endgroup$ – James Jan 16 at 4:24
  • $\begingroup$ 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? $\endgroup$ – McMath Jan 16 at 4:29


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?

  • $\begingroup$ To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx $\endgroup$ – McMath Jan 16 at 3:57
  • $\begingroup$ McMath: see my answer. $\endgroup$ – James Jan 16 at 4:04
  • $\begingroup$ @McMath, what if $m=1$ and $m=2?$ $\endgroup$ – lab bhattacharjee Jan 16 at 4:16
  • $\begingroup$ if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps??? $\endgroup$ – McMath Jan 16 at 4:38
  • $\begingroup$ @McMath Then $\cos x+\cos2x=?$ $\endgroup$ – lab bhattacharjee Jan 16 at 4:43

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

  • $\begingroup$ I've never seen this math formulas. Please explain using only basic trig identities. $\endgroup$ – McMath Jan 16 at 14:43

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