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So here is a trigonometric series.

$$\sin{55^\mathrm{o}}-\sin{19^\mathrm{o}}+\sin{53^\mathrm{o}}-\sin{17^\mathrm{o}}$$

Strange isn't it, and I have to calculate the total result of the series (without calculator). I don't think Maclaurin series will help me any way. Further I tried almost all trigonometric identities (as per my knowledge) but so far I had no clue. Probably I am missing some kind of identity. Anyone can help me in this?

Note: It's not a homework question.

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  • $\begingroup$ Hopefully arguments are degrees not radians. $\endgroup$
    – Jihad
    Dec 25, 2014 at 12:25
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    $\begingroup$ HINT: Use Prosthaphaeresis Formulas $\endgroup$
    – Mufasa
    Dec 25, 2014 at 12:30
  • $\begingroup$ @Jihad Yes. It's in degrees. $\endgroup$
    – Saharsh
    Dec 25, 2014 at 12:55

4 Answers 4

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Hint:

The key observation here is that $55+17=53+19$.

$$\sin{55}^\circ-\sin{19}^\circ+\sin{53}^\circ-\sin{17}^\circ=\color{#C00}{\sin55^\circ-\sin17^\circ}+\color{green}{\sin53^\circ-\sin19^\circ}.$$

Now use the identity

$$\sin a-\sin b=2\sin(\tfrac{a-b}2)\cos({\tfrac{a+b}2}).$$

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  • $\begingroup$ Thank you. This was the identity which I was missing. Silly me. $\endgroup$
    – Saharsh
    Dec 25, 2014 at 13:00
  • $\begingroup$ Glad I could help! $\endgroup$
    – Workaholic
    Dec 25, 2014 at 13:16
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You can use the following product-sum trig identity:

$$\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$$

$$\sin (a+b) + \sin (a-b) = 2 \sin a \cos b$$

So, $$(\sin 55^\circ - \sin 17^\circ) + (\sin 53^\circ - \sin 19^\circ)$$ $$(2 \cos 36^\circ \sin 19^\circ) + (2 \cos 36^\circ \sin 17^\circ)$$

$$2 \cos 36^\circ (\sin 19^\circ + \sin 17^\circ)$$

$$4 \cos 36^\circ \sin 18^\circ \cos 1^\circ$$


We can find $\sin 18^\circ$ using the following method, and use $\cos 2x = 1 - 2 \sin^2 x$ to find $\cos 36^\circ$.

$\sin 72^\circ = 2 \sin 36^\circ \cos 36^\circ$

$\cos 18^\circ = 4 \sin 18^\circ \cos 18^\circ (1 - 2 \sin^2 18^\circ)$

$1 = 4x(1 - 2x^2)$ where $x = \sin 18^\circ$

$8x^3 - 4x + 1 = 0$

We can factor using a few methods, I think the easiest is to observe that $x = 1/2$ is a root and do polynomial long division of $(2x-1)$:

$(2x - 1)(4x^2 + 2x - 1) = 0$ We can't have $x = \frac{1}{2}$ because that's $\sin 30^\circ$, not $\sin 18^\circ$.

Quadratic formula: $x = \frac{-2 \pm \sqrt{20}}{8} = \frac{1}{4} (-1 \pm \sqrt{5})$

The root is positive, so it's $\frac{1}{4} (-1 + \sqrt{5})$. From this follows $\cos 36^\circ = \frac{1}{4}(1 + \sqrt{5})$


So, the above answer reduces to just $\cos 1^\circ$. This doesn't have a nice formula, but it is algebraic!

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  • $\begingroup$ That's a neat way to find the exact values of sin(x) and cos(x) for not-so-nice values of x. I've never seen it before. Are there limitations to when this method can be used? $\endgroup$
    – Dasherman
    Dec 25, 2014 at 20:51
  • $\begingroup$ @Dasherman Yeah, there certainly are limitations. This case pretty much just happens to work out nicely. $\endgroup$
    – MT_
    Dec 26, 2014 at 5:32
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HINTS: The sum and difference of trig sines is to be used.See how 20 degree difference can be used as a common factor. 4 (Cos[36. Degree] Sin[18. Degree]) = 1 , must know the $ (\sqrt 5 \pm 1)/2 $ results.

Answer comes out as: $ \cos( 1^0)$

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  • $\begingroup$ I think you mean $(\sqrt{5} \pm 1)/4$ results $\endgroup$
    – MT_
    Dec 25, 2014 at 14:28
  • $\begingroup$ Yes, I mean generally one should know this Golden Ratio, sine, cos of multiples of 18 degrees etc., nice set of results. $\endgroup$
    – Narasimham
    Dec 25, 2014 at 15:32
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You need $\sin54^{\circ}-\sin18^{\circ}=1/2$.
If you know complex numbers, the following formula will be obvious: $$1+\cos72^{\circ}+\cos144^{\circ}+\cos216^{\circ}+\cos288^{\circ}=0$$ from which my first line follows.

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