A geometry problem by Jose Sacramento

Geometry Level 3


The answer is 2.

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

Michael Huang
Dec 2, 2016

Relevant wiki: Fundamental Trigonometric Identities - Problem Solving (Easy)

Section I. Equation A

By complementary angle identities , cos ⁡ ( π 22 ) = sin ⁡ ( π 2 − π 22 ) = sin ⁡ ( 5 π 11 ) cos ⁡ ( 7 π 12 ) = sin ⁡ ( 7 π 12 − π 2 ) = sin ⁡ ( π 12 ) \begin{array}{rl} {\color{#20A900}\cos\left(\dfrac{\pi}{22}\right)} &= \sin\left(\dfrac{\pi}{2} - \dfrac{\pi}{22}\right) = {\color{#20A900}\sin\left(\dfrac{5\pi}{11}\right)}\\ {\color{#3D99F6}\cos\left(\dfrac{7\pi}{12}\right)} &= \sin\left(\dfrac{7\pi}{12} - \dfrac{\pi}{2}\right) = {\color{#3D99F6}\sin\left(\dfrac{\pi}{12}\right)} \end{array} Applying the following Pythagorean identity sin ⁡ 2 ( θ ) + cos ⁡ 2 ( θ ) = 1 \sin^2\left(\theta\right) + \cos^2\left(\theta\right) = 1 gives A = cos ⁡ 2 ( 5 π 11 ) + cos ⁡ 2 ( π 22 ) + cos ⁡ 2 ( 7 π 12 ) + cos ⁡ 2 ( π 12 ) = 1 + 1 = 2 \begin{array}{rl} A = \cos^2\left(\dfrac{5\pi}{11}\right) + {\color{#20A900}\cos^2\left(\dfrac{\pi}{22}\right)} + {\color{#3D99F6}\cos^2\left(\dfrac{7\pi}{12}\right)} + \cos^2\left(\dfrac{\pi}{12}\right) = 1 + 1 = 2 \end{array}

Section II. Equation B

Since 14 π 9 , 13 π 9 ≥ π \dfrac{14\pi}{9}, \dfrac{13\pi}{9} \geq \pi , sin ⁡ < 0 \sin < 0 . In this case, express sin ⁡ ( 14 π 9 ) = − sin ⁡ ( 14 π 9 − π ) = − sin ⁡ ( 5 π 9 ) sin ⁡ ( 13 π 9 ) = − sin ⁡ ( 13 π 9 − π ) = − sin ⁡ ( 4 π 9 ) \begin{array}{rl} {\color{#D61F06}\sin\left(\dfrac{14\pi}{9}\right)} &= -\sin\left(\dfrac{14\pi}{9} - \pi\right) = {\color{#D61F06}-\sin\left(\dfrac{5\pi}{9}\right)}\\ {\color{#69047E}\sin\left(\dfrac{13\pi}{9}\right)} &= -\sin\left(\dfrac{13\pi}{9} - \pi\right) = {\color{#69047E}-\sin\left(\dfrac{4\pi}{9}\right)} \end{array} which yield B = ( sin ⁡ ( 4 π 9 ) + sin ⁡ ( 14 π 9 ) ) + ( sin ⁡ ( 5 π 9 ) + sin ⁡ ( 13 π 9 ) ) = 0 \begin{array}{rl} B &= \left(\sin\left(\dfrac{4\pi}{9}\right) + {\color{#D61F06}\sin\left(\dfrac{14\pi}{9}\right)}\right) + \left(\sin\left(\dfrac{5\pi}{9}\right) + {\color{#69047E}\sin\left(\dfrac{13\pi}{9}\right)}\right)\\ &= 0 \end{array}

Section III. Equation C

Observe that cos ⁡ ( 14 π 25 ) = − cos ⁡ ( 11 π 25 ) \begin{array}{rl} \cos\left(\dfrac{14\pi}{25}\right) &= -\cos\left(\dfrac{11\pi}{25}\right) \end{array} and cos ⁡ ( 15 π 26 ) = − sin ⁡ ( 15 π 26 − π ) = − sin ⁡ ( π 13 ) \begin{array}{rl} \cos\left(\dfrac{15\pi}{26}\right) &= -\sin\left(\dfrac{15\pi}{26} - \pi\right) = -\sin\left(\dfrac{\pi}{13}\right)\\ \end{array} Then, C = ( sin ⁡ ( π 13 ) + cos ⁡ ( 15 π 26 ) ) + ( cos ⁡ ( 11 π 25 ) + cos ⁡ ( 14 π 25 ) ) = 0 C = \left(\sin\left(\dfrac{\pi}{13}\right) + \cos\left(\dfrac{15\pi}{26}\right)\right) + \left(\cos\left(\dfrac{11\pi}{25}\right) +\cos\left(\dfrac{14\pi}{25}\right)\right) = 0

Section IV. Answer

Therefore, A + B + C = 2 A + B+ C = \boxed{2} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Very clear explanations for these sums :)

In this case, they only look scary, but are succumb to the basic properties.

Calvin Lin Staff - 4 years, 6 months ago

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