How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$

Question

Question: $ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $

I have partially solved this:-

$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$

$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$

$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$

$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$

$$ \sin(18^\circ) - \cos(36^\circ) $$

At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.

@user2838619, please, please, please for the sake of love don't do that when you have Latex available. :-( — Zain Patel, Jul 19 '15 at 13:54
@ajotatxe : See my latest edit to the question. ${}\qquad{}$ — Michael Hardy, Jul 19 '15 at 16:37
@ZainPatel : The braces aren't actually necessary. One can write ^\circ. Braces are needed for superscripts only when the supercript contains more than one item, as in $x^{35}$ (with two digits in the superscript). ${}\qquad{}$ — Michael Hardy, Jul 19 '15 at 23:44
@MichaelHardy, Ah, I figured that \circ would count as 4 items, each letter being counted as one item in the same way that each digit was in 35, thanks for that! :-) — Zain Patel, Jul 19 '15 at 23:46
Result is readily available from https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2?noredirect=1&lq=1 — lab bhattacharjee, Jun 18 '17 at 11:32

ajotatxe · Answer 1 · 2015-07-19T14:03:05.443

Angles that are multple of $18$ degrees occur in the regular pentagon with its diagonals and their trigonometric ratios are related to the golden ratio.

To find them, take an isosceles triangle with angles $A=36$ and $B=C=72$ degrees. Draw the bisector of $B$ that intersects $AC$ in a point $D$. Note that the triangles $ABC$ and $BCD$ are similar. Now let $1=AB$, $x=AD$. Note now that $ABD$ is also isosceles, and $AD=BD=BC=x$. The mentioned similarity yields $$\frac1x=\frac x{1-x}$$ which is precisely the equation that defines the golden ratio.

To find $\cos 36^\circ$ draw the altitude of the triangle $ABD$ from $D$, that bisects the side $AB$. Then $$\cos 36^\circ=\frac{1/2}{x}$$

Now, with the classical trigonometric identities, you can easily find closed expressions for the sine and the cosine of $18$, $36$, $54$ and $72$ degrees. For the angle of $18$ degrees, don't use half-angle formulae, but $$\sin 18^\circ=\sin(90^\circ-72^\circ)=\cos 72^\circ=\cos^2 36^\circ-\sin^2 36^\circ$$

Remark: Perhaps you are wandering what that pentagon had to do with all of this. If you choose a vertex $A$ of a regular pentagon and draw the two diagonals $AB$ and $AC$ from it, the triangle $ABC$ is similar to the one we have just used.

score 2 · Accepted Answer · answered Jul 19 '15 at 13:57

2

There is not fast but strightforward way. $18^\circ = \pi/10$; let $s=\sin{(\pi/10)}$ so, we should evaluate $$ a=\sin\frac\pi{10}-\cos\frac{\pi}{5}=2s^2 + s - 1. $$ We may express $\sin5x$ in terms of $\sin x$: $$ \sin5x=16\sin^5 x - 20\sin^3 x + 5\sin x $$ (i.e. from $\sin5x=\sin(4x + x), \sin4x=2\sin2x\cos2x$, or by de Moivre). For $x=\frac{\pi}{10}$ we have $\sin5x=1$: $$ 16s^5 - 20s^3 + 5s -1 =0, $$ or $$ (s-1)(-1+2s+4s^2)^2 = 0; $$ So, $$ 4s^2 + 2s - 1= 0=2a+1 \Longrightarrow a = -\frac12 $$ is answer.

answered Jul 19 '15 at 13:57

Michael Galuza

4,730

Could you explain how $ \sin\frac\pi{10}-\cos\frac{\pi}{5} $ is equal to $ 2s^2 + s - 1 $. – Abhishek Mhatre Jul 19 '15 at 14:12
@user2468338 $\cos 2x = 1 - 2\sin^2 x$ – Michael Galuza Jul 19 '15 at 14:16
How did you factor $ 16s^5 - 20s^3 + 5s - 1 $ – Abhishek Mhatre Jul 19 '15 at 14:19
@user2468338, actually, with Maple's help))) But $s=1$ is a root, and you should to factor $16s^4+16s^3-4s^2-4s+1$ – Michael Galuza Jul 19 '15 at 14:25
1

@user2468338 Since the quintic $16s^5-20s^3+5s-1$ is a polynomial with integer coefficients, we may use the rational root test to quickly determine that any rational roots must be powers of $\frac12$ up to $\frac{1}{16}$. Plugging $s=\frac{1}{2^n}$ we see that $s=1$ is the only rational root. After that, there are standard methods for factoring quartics. – David H Jul 19 '15 at 14:37

score 1 · Answer 3 · answered Jul 19 '15 at 13:54

You have done your question already. Note that $sin18^\circ=\frac{\sqrt{5}-1}{4}$ and $cos36^\circ=\frac{\sqrt{5}+1}{4}$.If you had done mulitiple angles in trigonometry you can easily calculate these.

For $sin18^\circ$ let, $A=18^\circ$. Then $5A=90^\circ$

$2A=90^\circ-3A$

$Sin2A=cos3A$. Use $Sin2A=2sinAcosA$ and $Cos3A=4cos^3A-3CosA.$

score 1 · Answer 4 · answered Jul 19 '15 at 14:42

$\sin(5\cdot78^\circ)=\sin(360^\circ+30^\circ)=\sin30^\circ$

$\sin\{5(-66^\circ)\}=\sin(-360^\circ+30^\circ)=\sin30^\circ$

If $\sin5x=\sin30^\circ\implies5x=n180^\circ+(-1)^n30^\circ$ where $n$ is any integer

$\implies x=n72^\circ+6^\circ$ where $n\equiv-2,-1,0,1,2\pmod5$

Again, $\sin5x=16\sin^5x-20\sin^3x+5\sin x$

So, the roots of $\displaystyle16\sin^5x-20\sin^3x+5\sin x=\dfrac12$ are $\sin\left(n72^\circ+6^\circ\right)$ where $n\equiv-2,-1,0,1,2\pmod5$

Using Vieta's formula, $\displaystyle\sum_{n=-2}^2\sin\left(n72^\circ+6^\circ\right)=0$

$n=-2\implies$ $-2\cdot72^\circ+6^\circ=-138^\circ\implies\sin(-138^\circ)=-\sin(138^\circ)=-\sin(180^\circ-42^\circ)=-\sin42^\circ$

$n=-1\implies$ $-1\cdot72^\circ+6^\circ=-66^\circ\implies\sin(-66^\circ)=-\sin66^\circ$

$n=0\implies ?$

$n=1\implies ?$

and $n=2\implies\sin\left(2\cdot72^\circ+6^\circ\right)=\sin150^\circ=\dfrac12$

Do you see the destination?

score 0 · Answer 5 · edited Apr 13 '17 at 12:19

As $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$

$S=\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ $

$=\sin(-138^\circ)+\sin(-66^\circ)+\sin6°+\sin78^\circ $

Observe that the angles are in Arithmetic Progression with common difference $=72^\circ$

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

and Werner Formula: $2\sin A\sin B=\cos(A-B)-\cos(A+B),$

and finally Prosthaphaeresis Formula, $\cos C+\cos D=2\cos\dfrac{C+D}2\cos\dfrac{C-D}2$

$2\sin36^\circ\cdot S=-2\cos54^\circ\cos60^\circ\implies S=-\dfrac12$

How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$

5 Answers5