Sum of cosines and product of sines

Geometry Level 3

cos ⁡ ( 1 2 ∘ ) + cos ⁡ ( 8 4 ∘ ) + cos ⁡ ( 15 6 ∘ ) + cos ⁡ ( 13 2 ∘ ) sin ⁡ ( 6 ∘ ) sin ⁡ ( 4 2 ∘ ) sin ⁡ ( 6 6 ∘ ) sin ⁡ ( 7 8 ∘ ) = ? \large{\dfrac{\cos(12^{\circ})+\cos(84^{\circ})+\cos(156^{\circ})+\cos(132^{\circ})}{\sin(6^{\circ}) \sin(42^{\circ}) \sin(66^{\circ}) \sin(78^{\circ})}= \ ? }


The answer is -8.

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Tanishq Varshney by Tanishq Varshney , 23, India

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2 solutions

Jessica Wang
Jul 16, 2015

We can divide this question into two parts, i.e. the numerator and the denominator.

[ 1 ] . cos ⁡ 1 2 ∘ + cos ⁡ 8 4 ∘ + cos ⁡ 15 6 ∘ + cos ⁡ 13 2 ∘ [1].\: \cos 12^{\circ}+\cos84^{\circ}+\cos156^{\circ}+\cos132^{\circ}

= cos ⁡ 1 2 ∘ − cos ⁡ 2 4 ∘ − cos ⁡ 4 8 ∘ − cos ⁡ 9 6 ∘ =\cos12^{\circ}-\cos24^{\circ}-\cos48^{\circ}-\cos96^{\circ}

= sin ⁡ 7 8 ∘ − sin ⁡ 6 6 ∘ − sin ⁡ 4 2 ∘ + sin ⁡ 6 ∘ =\sin78^{\circ}-\sin66^{\circ}-\sin42^{\circ}+\sin6^{\circ}

= ( sin ⁡ 7 8 ∘ − sin ⁡ 4 2 ∘ ) − ( sin ⁡ 6 6 ∘ − sin ⁡ 6 ∘ ) =(\sin78^{\circ}-\sin42^{\circ})-(\sin66^{\circ}-\sin6^{\circ})

= 2 cos ⁡ 6 0 ∘ sin ⁡ 1 8 ∘ − 2 cos ⁡ 3 6 ∘ sin ⁡ 3 0 ∘ =2\cos60^{\circ}\sin18^{\circ}-2\cos36^{\circ}\sin30^{\circ}

= sin ⁡ 1 8 ∘ − cos ⁡ 3 6 ∘ =\sin18^{\circ}-\cos36^{\circ}

= sin ⁡ 1 8 ∘ − sin ⁡ 5 4 ∘ =\sin18^{\circ}-\sin54^{\circ}

= − 2 cos ⁡ 3 6 ∘ sin ⁡ 1 8 ∘ =-2\cos36^{\circ}\sin18^{\circ}

= − 2 cos ⁡ 3 6 ∘ sin ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ =-\frac{2\cos36^{\circ}\sin18^{\circ}\cos18^{\circ}}{\cos18^{\circ}}

= − cos ⁡ 3 6 ∘ sin ⁡ 3 6 ∘ cos ⁡ 1 8 ∘ =-\frac{\cos36^{\circ}\sin36^{\circ}}{\cos18^{\circ}}

= − 0.5 sin ⁡ 7 2 ∘ cos ⁡ 1 8 ∘ =-\frac{0.5\sin72^{\circ}}{\cos18^{\circ}}

= − 0.5 cos ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ =-\frac{0.5\cos18^{\circ}}{\cos18^{\circ}}

= − 0.5 =-0.5 ;

[ 2 ] . sin ⁡ 6 ∘ sin ⁡ 4 2 ∘ sin ⁡ 6 6 ∘ sin ⁡ 7 8 ∘ [2].\: \sin6^{\circ}\sin42^{\circ}\sin66^{\circ}\sin78^{\circ}

= sin ⁡ 6 ∘ cos ⁡ 4 8 ∘ cos ⁡ 2 4 ∘ cos ⁡ 1 2 ∘ =\sin6^{\circ}\cos48^{\circ}\cos24^{\circ}\cos12^{\circ}

= 2 cos ⁡ 6 ∘ sin ⁡ 6 ∘ cos ⁡ 1 2 ∘ cos ⁡ 2 4 ∘ cos ⁡ 4 8 ∘ 2 cos ⁡ 6 ∘ =\frac{2\cos6^{\circ}\sin6^{\circ}\cos12^{\circ}\cos24^{\circ}\cos48^{\circ}}{2\cos6^{\circ}}

= sin ⁡ 1 2 ∘ cos ⁡ 1 2 ∘ cos ⁡ 2 4 ∘ cos ⁡ 4 8 ∘ 2 cos ⁡ 6 ∘ =\frac{\sin12^{\circ}\cos12^{\circ}\cos24^{\circ}\cos48^{\circ}}{2\cos6^{\circ}}

= 2 sin ⁡ 1 2 ∘ cos ⁡ 1 2 ∘ cos ⁡ 2 4 ∘ cos ⁡ 4 8 ∘ 4 cos ⁡ 6 ∘ =\frac{2\sin12^{\circ}\cos12^{\circ}\cos24^{\circ}\cos48^{\circ}}{4\cos6^{\circ}}

= sin ⁡ 2 4 ∘ cos ⁡ 2 4 ∘ cos ⁡ 4 8 ∘ 4 cos ⁡ 6 ∘ =\frac{\sin24^{\circ}\cos24^{\circ}\cos48^{\circ}}{4\cos6^{\circ}}

= 2 sin ⁡ 2 4 ∘ cos ⁡ 2 4 ∘ cos ⁡ 4 8 ∘ 8 cos ⁡ 6 ∘ =\frac{2\sin24^{\circ}\cos24^{\circ}\cos48^{\circ}}{8\cos6^{\circ}}

= sin ⁡ 4 8 ∘ cos ⁡ 4 8 ∘ 8 cos ⁡ 6 ∘ =\frac{\sin48^{\circ}\cos48^{\circ}}{8\cos6^{\circ}}

= 2 sin ⁡ 4 8 ∘ cos ⁡ 4 8 ∘ 16 cos ⁡ 6 ∘ =\frac{2\sin48^{\circ}\cos48^{\circ}}{16\cos6^{\circ}}

= sin ⁡ 9 6 ∘ 16 cos ⁡ 6 ∘ =\frac{\sin96^{\circ}}{16\cos6^{\circ}}

= cos ⁡ 6 ∘ 16 cos ⁡ 6 ∘ =\frac{\cos6^{\circ}}{16\cos6^{\circ}}

= 1 16 =\frac{1}{16} .

∴ ( − 0.5 ) ÷ ( 1 16 ) = − 8 . \therefore (-0.5)\div (\frac{1}{16})=\boxed{-8}.

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Moderator note:

The denominator has a somewhat recognizable form of doubling the angles, which would suggest such an approach.

What is the form of the numerator that would make use suspect that it is so "factorable"?

@Jessica Wang , you might like my alternative solution. Enjoy!

Pi Han Goh - 5 years, 6 months ago

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Niccceeeeee! Thank you for your information! @Pi Han Goh

Jessica Wang - 5 years, 6 months ago

Cos 132 is same as cos 228 which makes the numerator sum of cosines of angles in AP which can be caculated by
(Sin nd/2 ÷sin d/2 ) * cos ( a + (n - 1)d /2) Where a is first angle n is no. of angles and d is common difference between the two angles.

Shaurya Gomber - 5 years, 5 months ago
Pi Han Goh
Nov 6, 2015

Synopsis : I will compute the values of the numerator and denominator separately.

Numerator: Apply the identity cos ⁡ ( π n ) + cos ⁡ ( ( 3 π n ) + cos ⁡ ( ( 5 π n ) + … + cos ⁡ ( ( n − 2 ) π n ) = 1 2 \cos\left( \frac\pi{n} \right) + \cos(\left( \frac{3\pi}n\right) + \cos(\left( \frac{5\pi}n\right) + \ldots + \cos\left( \frac{(n-2)\pi}n\right) = \dfrac12 for positive odd integer n n . See the proof here .

Denominator: Apply the Trigonometric Equations - Triple Angle Formula : 4 sin ⁡ ( x ) sin ⁡ ( 6 0 ∘ + x ) sin ⁡ ( 6 0 ∘ − x ) = sin ⁡ ( 3 x ) 4\sin(x) \sin(60^\circ+ x) \sin(60^\circ - x) = \sin(3x) .

Numerator:

Note that 12 180 , 84 180 , 156 180 , 132 180 \frac{12}{180}, \frac{84}{180}, \frac{156}{180} ,\frac{132}{180} reduced to the form 1 15 , 7 15 , 13 15 , 11 15 \frac1{15}, \frac{7}{15} ,\frac{13}{15}, \frac{11}{15} . And 1 , 7 , 13 , 11 1,7,13,11 are all positive integers less than 15 that are coprime to 15. This motivates me to use the identity:

cos ⁡ ( π 15 ) + cos ⁡ ( 3 π 15 ) + cos ⁡ ( 5 π 15 ) + cos ⁡ ( 7 π 15 ) + cos ⁡ ( 9 π 15 ) + cos ⁡ ( 11 π 15 ) + cos ⁡ ( 13 π 15 ) = 1 2 \cos\left( \frac{\pi}{15}\right) + \cos\left( \frac{3\pi}{15}\right) + \cos\left( \frac{5\pi}{15}\right) + \cos\left( \frac{7\pi}{15}\right) + \cos\left( \frac{9\pi}{15}\right) +\cos\left( \frac{11\pi}{15}\right) + \cos\left( \frac{13\pi}{15}\right) = \frac12

Similarly, cos ⁡ ( π 5 ) + cos ⁡ ( 3 π 5 ) = 1 2 \cos\left( \frac{\pi}{5}\right) + \cos\left( \frac{3\pi}{5}\right) = \frac12 .

Substituting everything and simplify, the numerator is equal to − 1 2 -\dfrac12 .

Denominator:

We apply the identity 4 sin ⁡ ( x ) sin ⁡ ( 6 0 ∘ + x ) sin ⁡ ( 6 0 ∘ − x ) = sin ⁡ ( 3 x ) 4\sin(x) \sin(60^\circ+ x) \sin(60^\circ - x) = \sin(3x) . Substitute x = 6 ∘ x = 6^\circ and x = 1 8 ∘ x = 18^\circ , we get

4 sin ⁡ ( 6 ∘ ) sin ⁡ ( 6 6 ∘ ) sin ⁡ ( 5 4 ∘ ) = sin ⁡ ( 1 8 ∘ ) 4 sin ⁡ ( 1 8 ∘ ) sin ⁡ ( 7 8 ∘ ) sin ⁡ ( 4 2 ∘ ) = sin ⁡ ( 5 4 ∘ ) 4\sin(6^\circ) \sin(66^\circ) \sin(54^\circ) = \sin(18^\circ) \\ 4\sin(18^\circ) \sin(78^\circ) \sin(42^\circ) = \sin(54^\circ)

Multiplying both sides and simplify, we obtain the denominator with a value of 1 16 \dfrac1{16} .

Taking their ratio, we get − 8 \boxed{-8} .

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