Minimario's trigonometric expression

First, see that: $\cos x = \dfrac{\sin 2x}{2\sin x}$

Now, we can rewrite the expression as:

$\dfrac{\sin 24°}{2\sin 12°}\times\dfrac{\sin 48°}{2\sin 24°}\times\dfrac{\sin 96º}{2\sin 48°}\times\dfrac{\sin192°}{2\sin96°}$ $=\dfrac{\sin 192°}{16\sin 12º}=\dfrac{-\sin 12°}{16\sin 12°}=\frac{-1}{16}$ Then, $a=1$ and $b=16\ \therefore a+b=17$

Call the above $S$ .

We notice that those degrees are increasing by double so we decide to use the sin double angle formula which is

$\sin 2\theta = 2\sin\theta\cos\theta$ or $\frac{1}{2}\sin 2\theta = \sin\theta\cos\theta$

Multiply $S$ by $\sin12^\circ$ to get

$S\sin 12^\circ = \sin12^\circ \times \cos12^\circ \times \cos24^\circ \times \cos48^\circ \times \cos96^\circ$

$S\sin 12^\circ = \frac{1}{2}\sin24^\circ \times \cos24^\circ \times \cos48^\circ \times \cos96^\circ$

$S\sin 12^\circ = \frac{1}{4}\sin48^\circ \times \cos48^\circ \times \cos96^\circ$

$S\sin 12^\circ = \frac{1}{8}\sin96^\circ \times \cos96^\circ$

$S\sin 12^\circ = \frac{1}{16}\sin192^\circ$

$\sin12^\circ = -\sin192^\circ$

$S = -\frac{1}{16}$

$a+b=17$

We know that $\cos (x\pm y)=\cos x \cos y \mp \sin x \sin y$ Adding both identities and rearranging terms we can get $\cos x \cos y = \frac{1}{2} \left[ \cos (x+y)+ \cos (x-y) \right]$ Now, let's call $P=\cos 12° \cdot \cos 24° \cdot \cos 48° \cdot \cos 96°$ , but we use this identity pairing up the 12 with the 48, and 24 with the 96. We obtain $P=\frac{1}{2} \left( \cos 60° + \cos 36° \right) \times \frac{1}{2} \left( \cos 120° + cos 72° \right)$ We are implicitly using $\cos x = \cos(-x)$ . We can simplify $P$ a bit further, since we know $\cos 60° = \frac{1}{2}$ and $\cos 120°=-\frac{1}{2}$ , that is $P=\frac{1}{4}\left( \frac{1}{2} + \cos 36° \right) \left( -\frac{1}{2} + \cos 72°\right)$ All we need to do now, is to calculate the $\cos 36°$ and $\cos 72°$ , which is easy. I'll leave that detail until the end. For now, i'll use that $\cos 36° = \frac{1+\sqrt{5}}{4}$ and $\cos 72° = \frac{1+\sqrt{5}}{4}$ . Substituting this into the expression for $P$ we get $P=\frac{1}{4}\left( \frac{1}{2} + \frac{1+\sqrt{5}}{4} \right) \left( -\frac{1}{2} + \frac{1+\sqrt{5}}{4}\right)$ Now, doing a bit of algebra we get $P = \frac{1}{4} \left( \frac{3+\sqrt{5}}{4} \right) \left( \frac{-3+\sqrt{5}}{4} \right)$ which leads to $P=\frac{1}{4^3}\left(-9 + 5\right)$ so that $\boxed{P=-\frac{1}{16}}$ That way, the answer for this problema is $\boxed{a+b = 17}$ .

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Consider an isosceles triangle $OAB$ with $OA = OB$ and $\angle AOB = 36°$ . Let be $C$ a point in segment $OB$ such tha $AC$ is the \angle bisector. It is easy to see $\angle AOC = \angle CAB = 36°\quad \text{and} \quad \angle CBA = \angle ACB = 72°$ which proves that triangles $OAB$ and $ABC$ are similar. Without losing generality, let $OA=OB=1$ and $AB=AC=OC=x$ , because triangle $OAC$ is isosceles too. By this similarity, $\frac{BC}{AB}=\frac{AB}{OA}$ which is $\frac{1-x}{x}=\frac{x}{1}$ Solving for $x$ and considering it must be positive, the only solution posible is $x=\frac{-1+\sqrt{5}}{2}$ . Let it be $AD$ the height of triangle $OAB$ , its obvious that $D$ is the midpoint of $CB$ . Then, $OD= x + \frac{1-x}{2}=\frac{1+x}{2}=\frac{1+\sqrt{5}}{4}$ But notice that $\cos36° = \frac{OD}{OA}=\frac{OD}{1}$ , that way we get $\cos 36° = \frac{1+\sqrt{5}}{4}$ Furthermore, since $\cos 2x = 2\cos^2 x - 1$ doing some algebra we observe that $\cos 72° = \frac{-1+\sqrt{5}}{4}.$

To begin with, note that

• $\sin2\theta=2\sin\theta \cos\theta$

• $\sin(180^{\circ}+\theta)=-\sin\theta$

Now, let us multiply the entire expression with $\frac{1}{2\sin12^{\circ}} \cdot 2\sin12^{\circ}$ , which is essentially $1$ , meaning that it will cause no change to the expression. Therefore, we have

$\frac{1}{2\sin12^{\circ}} \cdot (2\sin12^{\circ}) \cdot (\cos12^{\circ}) \cdot (\cos24^{\circ}) \cdot (\cos48^{\circ}) \cdot (\cos96^{\circ})$

Now, since $2\sin12^{\circ}\cos12^{\circ}=\sin24^{\circ}$ , we may substitute that in the expression, leaving us with

$\frac{1}{2\sin12^{\circ}} \cdot (\sin24^{\circ}) \cdot (\cos24^{\circ}) \cdot (\cos48^{\circ}) \cdot (\cos96^{\circ})$

Now, let us multiply the expression with $\frac{1}{2}\cdot2$ , which is again $1$ and does not change anything. Thus

$\frac{1}{4\sin12^{\circ}} \cdot (2\sin24^{\circ}) \cdot (\cos24^{\circ}) \cdot (\cos48^{\circ}) \cdot (\cos96^{\circ})$

... and again, seeing that $2\sin24^{\circ}\cos24^{\circ}=\sin48^{\circ}$ , we may substitute it to the expression, leaving us with

$\frac{1}{4\sin12^{\circ}} \cdot (\sin48^{\circ}) \cdot (\cos48^{\circ}) \cdot (\cos96^{\circ})$

At this point, we repeat the second step (multiplying with $\frac{1}{2}\cdot2$ ) $2$ more times, so that we get

$\frac{1}{4\sin12^{\circ}} \cdot (\sin48^{\circ}) \cdot (\cos48^{\circ}) \cdot (\cos96^{\circ})=$

$=\frac{1}{8\sin12^{\circ}} \cdot (\sin96^{\circ}) \cdot (\cos96^{\circ})= \frac{1}{16\sin12^{\circ}} \cdot (\sin192^{\circ})$ .

Considering $\sin192^{\circ}=\sin(180^{\circ}+12^{\circ}=-\sin12^{\circ}$ , we conclude that

$\frac{1}{16\sin12^{\circ}} \cdot (\sin192^{\circ})=\frac{1}{16\sin12^{\circ}} \cdot (-\sin12^{\circ})=-\frac{1}{16}$

From the text of the problem we can see that $-\frac{a}{b}=-\frac{1}{16}$ , so $a=1$ and $b=16$ , meaning $a+b=17$ .

By double angle formula, we have $\sin 24^\circ = 2\sin 12^\circ \cos 12^\circ \implies \cos 12^\circ =\frac{ \sin 24^\circ}{2\sin 12^\circ}$ $\sin 48^\circ = 2\sin 24^\circ \cos 24^\circ \implies \cos 24^\circ =\frac{ \sin 48^\circ}{2\sin 24^\circ}$ $\sin 96^\circ = 2\sin 48^\circ \cos 48^\circ \implies \cos 48^\circ =\frac{ \sin 96^\circ}{2\sin 48^\circ}$ $\sin 192^\circ = 2\sin 96^\circ \cos 96^\circ \implies \cos 96^\circ =\frac{ \sin 192^\circ}{2\sin 96^\circ}$ So multiplying these 4 gives $\frac{\sin 192^\circ}{16\sin 12^\circ} = \frac{-\sin 12^\circ}{16\sin 12^\circ} = - \frac 1{16}\implies a + b = 17.$

We first need to convert all there terms into sin so we 1st multiply ans divide by 2sin (12)

2 sin (12)×cos(12)×cos (24)×cos (48)×cos (96) /2 sin (12) If we see 1st two terms they are of form 2sinx cos x which is equal to sin (2x) we get

Sin (24)×cos (24)×cos (48)×cos (96)/2sin (12) Now we need to multiply and divide by 2

2sin (24)×cos (24)×cos (48)×cos (96)/4 sin (12)

Sin (48)×cos (48)×cos (96)/4sin (12)

Again by multiplying and dividing by 2 at lat we get

Sin (192)/16sin (12)

We know sin (180+x)=-sin (x)

So we write -sin (12)/16sin (12)

We get -1/16 so a+b=17

For these series of form Cos (a)×cos (2a)×cos (4a)×.............×cos (2^(n-1) a)

We get result sin (2^n a)/2^n sin (a)

We have $(\cos12^\circ ) \times (\cos24^\circ) \times (\cos48^\circ) \times (\cos96^\circ) = (\cos12^\circ) \times (\cos48^\circ) \times (\cos24^\circ) \times (\cos96^\circ)$

$= \frac{\cos60^\circ + \cos36^\circ}{2} \times \frac{\cos120^\circ + \cos72^\circ}{2}$ $= \frac{1}{4} \left( \frac{1}{2} + \cos36^\circ \right) \left(\frac{-1}{2} + \cos72^\circ \right)$ $= \frac{1}{4} \left( \frac{-1}{4} -\frac{1}{2}\cos36^\circ +\frac{1}{2}\cos72^\circ +\cos36^\circ\cos72^\circ \right)$ $= \frac{1}{4} \left( \frac{-1}{4} -\frac{1}{2}\cos36^\circ +\frac{1}{2}\cos72^\circ +\frac{\cos108^\circ+\cos36^\circ}{2} \right)$ $= \frac{1}{4} \left( \frac{-1}{4} -\frac{1}{2}\cos36^\circ +\frac{1}{2}\cos72^\circ +\frac{-\cos72^\circ+\cos36^\circ}{2} \right)$ $= \frac{1}{4} \left( \frac{-1}{4} \right)$ $= \frac{-1}{16}$

Thus $a=1, b=16$ and $a+b=17$

Multiply two sides of the equation with $sin\ 12^{∘}$ we have:

$sin\ 12^{∘}$ x $cos\ 12^{∘}$ x $cos\ 24^{∘}$ x $cos\ 48^{∘}$ x $cos\ 96^{∘} = - \frac{a}{b}$ x $sin\ 12^{∘}$ $<=> \frac{1}{16}$ x $sin\ 192^{∘} = - \frac{a}{b}$ x $sin\ 12^{∘}$

$<=> - \frac{1}{16}$ x $sin\ 12^{∘} = - \frac{a}{b}$ x $sin\ 12^{∘}$

$<=> \frac{1}{16} = \frac{a}{b}$

$=> a + b = 17$

So we have 17 as the result.

Multiplying and dividing by cos 72,,we get cos 12 * cos 48* cos 72 cos 24 cos 96/cos 72. IN THE NEXT STEP,,we can write it as cos12 cos(60-12) cos(60+12) cos 24 cos 96/cos 72. we can simplify it as cos(3X12)cos 24 cos 96/cos 72. =cos 36 cos 24 cos 96/cos 72 cos (60-36) cos 36 cos(60+36)/cos 72 =cos(3X36)/cos72 cos 108/cos72 -cos18/sin 18 -cot18 =-13/4. THE FORMULA IS COS(60-A) COSA COS(60+A) = COS 3A SORRY FRIENDS FOR NOT WRITTING MATHEMATICALLY...

cos 12. cos 24. cos 48. cos 96= [1 / (2 sin 12) ] ( 2. sin 12. cos 12 ). cos 24. cos 48. cos 96 = [1/ (2 sin 12 )] ( sin 24 ) cos 24. cos 48. cos 96 = [1 / (4 sin 12 )] ( 2. sin 24. cos 24 ). cos 48. cos 96 = [ 1 /(4 sin 12 )] ( sin 48 ).cos 48. cos 96 = [ 1 / (8 sin 12 )].( 2. sin 48. cos 48 ). cos 96 = [1 / (8 sin 12 )].( sin 96 ). cos 96 = [ 1 / (16 sin 12)]. ( 2. sin 96. cos 96 ) = [1 / (16 sin 12)]. sin 192 = [1 / (16 sin 12 )]. sin [ 180 + 12 ] = [1 / (16 sin 12 )].( - sin 12 )= ( - 1 / 16 ).=>1+16=17

We will use :

[ 1 ] 2* sin x. cos x = sin 2x

[ 2 ] sin ( 180 + x ) = - sin x. ......................................…

= cos 12. cos 24. cos 48. cos 96

= ( 1 / 2*sin 12 ) ( 2. sin 12. cos 12 ). cos 24. cos 48. cos 96

= ( 1/ 2*sin 12 ) ( sin 24 ) cos 24. cos 48. cos 96

= ( 1 / 4*sin 12 ) ( 2. sin 24. cos 24 ). cos 48. cos 96

= ( 1 /4*sin 12 ) ( sin 48 ).cos 48. cos 96

= ( 1 / 8*sin 12 ).( 2. sin 48. cos 48 ). cos 96

= ( 1 / 8*sin 12 ).( sin 96 ). cos 96

= ( 1 / 16*sin 12 ). ( 2. sin 96. cos 96 )

= ( 1 / 16* sin 12 ). sin 192

= ( 1 / 16*sin 12 ). sin [ 180 + 12 ]

= ( 1 / 16*sin 12 ).( - sin 12 )

= ( - 1 / 16 ). ............................Ans.

The given sum can be simplified as $\cos \theta$ $\cos2 \theta$ $\cos4 \theta$ $\cos8 \theta$ where $\theta$ = $12^{o}$ .

Now, multiplying the numerator and denominator by 2 $\sin \theta$ we get [ $\sin2 \theta$ $\cos2 \theta$ $\cos4 \theta$ $\cos8 \theta$ ] / 2 $\sin \theta$ .

[using $\sin2 \theta$ = 2 $\sin \theta$ $\cos \theta$ ]

Multiplying the numerator and denominator by 2 we get $\sin4 \theta$ $\cos4 \theta$ $\cos8 \theta$ ] / 4 $\sin \theta$ .

Continuing like this, we get $\sin16 \theta$ / 16 $\sin \theta$

16 $\theta$ =16x $12^{o}$ = $180^{o}$ + $12^{o}$

Hence, $\sin16 \theta$ = - $\sin \theta$

Thus the fraction becomes - $\frac{1}{16}$

Therefore, a+b=17