Trigonometry Challenge 4

Geometry Level 3

tan ⁡ 11 π 30 − tan ⁡ π 30 sin ⁡ 11 π 15 − sin ⁡ π 15 = ? \large \frac{\tan{\frac{11\pi}{30}}-\tan{\frac{\pi}{30}}}{\sin{\frac{11\pi}{15}}-\sin{\frac{\pi}{15}}}=\boxed{?}


The answer is 4.

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Noel Lo by Noel Lo , 24, Singapore

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4 solutions

Chew-Seong Cheong
Jul 25, 2017

X = tan ⁡ 11 π 30 − tan ⁡ π 30 sin ⁡ 11 π 15 − sin ⁡ π 15 Note: tan ⁡ ( π − x ) = − tan ⁡ x = − tan ⁡ 19 π 30 − tan ⁡ π 30 sin ⁡ ( 2 π 5 + π 3 ) − sin ⁡ ( 2 π 5 − π 3 ) = tan ⁡ π 3 − ( tan ⁡ 10 π 30 + tan ⁡ 19 π 30 + tan ⁡ π 30 ) 2 cos ⁡ 2 π 5 sin ⁡ π 3 If A + B + C = π ⟹ tan ⁡ A + tan ⁡ B + tan ⁡ C = tan ⁡ A tan ⁡ B tan ⁡ C = tan ⁡ π 3 − tan ⁡ π 3 tan ⁡ 19 π 30 tan ⁡ π 30 3 cos ⁡ 2 π 5 = 1 − sin ⁡ 19 π 30 sin ⁡ π 30 cos ⁡ 19 π 30 cos ⁡ π 30 cos ⁡ 2 π 5 = cos ⁡ 19 π 30 cos ⁡ π 30 − sin ⁡ 19 π 30 sin ⁡ π 30 cos ⁡ 2 π 5 cos ⁡ 19 π 30 cos ⁡ π 30 Note: cos ⁡ ( A + B ) = cos ⁡ A cos ⁡ B − sin ⁡ A sin ⁡ B = cos ⁡ 2 π 3 1 2 cos ⁡ 2 π 5 ( cos ⁡ 3 π 5 + cos ⁡ 2 π 3 ) Note: cos ⁡ A cos ⁡ B = 1 2 ( cos ⁡ ( A − B ) + cos ⁡ ( A + B ) ) = − 1 cos ⁡ 2 π 5 ( − cos ⁡ 2 π 5 − 1 2 ) Note: cos ⁡ ( π − x ) = − cos ⁡ x = 2 2 cos ⁡ 2 2 π 5 + cos ⁡ 2 π 5 Note: cos ⁡ 2 x = 2 cos ⁡ 2 x − 1 = 2 cos ⁡ 4 π 5 + cos ⁡ 2 π 5 + 1 Note: ∑ k = 1 n cos ⁡ 2 k π 2 n + 1 = − 1 2 = 2 − 1 2 + 1 = 4 \begin{aligned} X & = \frac {{\color{#3D99F6}\tan \frac{11\pi}{30}}- \tan \frac \pi{30}}{\sin \frac{11\pi}{15}- \sin \frac \pi{15}} & \small \color{#3D99F6} \text{Note: } \tan (\pi - x) = - \tan x \\ & = \frac {{\color{#3D99F6}-\tan \frac{19\pi}{30}}- \tan \frac \pi{30}}{\sin \left(\frac{2\pi}5 + \frac \pi 3\right) - \sin \left(\frac{2\pi}5 - \frac \pi 3\right)} \\ & = \frac {\tan \frac \pi 3 - \left({\color{#3D99F6}\tan \frac{10\pi}{30} +\tan \frac{19\pi}{30} + \tan \frac \pi{30}}\right)}{2 \cos \frac{2\pi}5 \sin \frac \pi 3} & \small \color{#3D99F6} \text{If }A+B+C = \pi \implies \tan A+\tan B + \tan C = \tan A\tan B\tan C \\ & = \frac {\tan \frac \pi 3 - \color{#3D99F6}\tan \frac \pi 3 \tan \frac{19\pi}{30}\tan \frac \pi{30}}{\sqrt 3 \cos \frac{2\pi}5} \\ & = \frac {1 - \frac {\sin \frac{19\pi}{30}\sin \frac \pi{30}}{\cos \frac{19\pi}{30}\cos \frac \pi{30}}}{\cos \frac{2\pi}5} \\ & = \frac {\color{#3D99F6}\cos \frac{19\pi}{30}\cos \frac \pi{30} - \sin \frac{19\pi}{30}\sin \frac \pi{30}}{\cos \frac{2\pi}5\color{#D61F06}\cos \frac{19\pi}{30}\cos \frac \pi{30}} & \small \color{#3D99F6} \text{Note: } \cos (A+B) = \cos A \cos B - \sin A \sin B \\ & = \frac {\color{#3D99F6}\cos \frac {2\pi} 3}{\color{#D61F06}\frac 12 {\color{#333333} \cos \frac{2\pi}5} \left(\cos \frac{3\pi}5 + \cos \frac {2\pi}3\right)} & \small \color{#D61F06} \text{Note: } \cos A \cos B = \frac 12 \left(\cos (A-B) + \cos (A+B)\right) \\ & = \frac {-1}{\cos \frac{2\pi}5 \left({\color{#3D99F6}- \cos \frac{2\pi}5} - \frac 12 \right)} & \small \color{#3D99F6} \text{Note: } \cos (\pi - x) = - \cos x \\ & = \frac 2{{\color{#3D99F6}2\cos^2 \frac{2\pi}5} + \cos \frac{2\pi}5} & \small \color{#3D99F6} \text{Note: } \cos 2x = 2\cos^2 x - 1 \\ & = \frac 2{{\color{#3D99F6}\cos \frac{4\pi}5 + \cos \frac{2\pi}5} + 1} & \small \color{#3D99F6} \text{Note: } \sum_{k=1}^n \cos \frac {2k\pi}{2n+1} = - \frac 12 \\ & = \frac 2{{\color{#3D99F6}-\frac 12} + 1} \\ & = \boxed{4} \end{aligned}

Note: ∑ k = 0 n − 1 cos ⁡ ( 2 k π 2 n + 1 ) = − 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k\pi}{2n+1} \right) = -\frac 12 (see proof here ).

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How is that ∑ k = 1 n cos ⁡ 2 n π 2 n + 1 = − 1 2 \sum_{k=1}^n \cos \frac {2n\pi}{2n+1} = - \frac 12 ?

Kelvin Hong - 3 years, 10 months ago

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The following is the proof for ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 π ) = 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 . Similar proof for ∑ k = 0 n − 1 cos ⁡ ( 2 k 2 n + 1 π ) = − 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k}{2n+1}\pi \right) = - \frac 12 .

S = ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 π ) = ℜ { ∑ k = 0 n − 1 e 2 k + 1 2 n + 1 π i } = ℜ { e π i 2 n + 1 ∑ k = 0 n − 1 e 2 k π i 2 n + 1 } = ℜ { e π i 2 n + 1 ( 1 − e 2 n π i 2 n + 1 1 − e 2 π i 2 n + 1 ) } = ℜ { e π i 2 n + 1 − e π i 1 − e 2 π i 2 n + 1 } = ℜ { e π i 2 n + 1 + 1 ( 1 + e π i 2 n + 1 ) ( 1 − e π i 2 n + 1 ) } = ℜ { 1 1 − e π i 2 n + 1 } = ℜ { 1 1 − cos ⁡ π i 2 n + 1 − i sin ⁡ π i 2 n + 1 } = ℜ { 1 − cos ⁡ π i 2 n + 1 + i sin ⁡ π i 2 n + 1 ( 1 − cos ⁡ π i 2 n + 1 ) 2 + sin ⁡ 2 π i 2 n + 1 } = ℜ { 1 − cos ⁡ π i 2 n + 1 + i sin ⁡ π i 2 n + 1 2 − 2 cos ⁡ π i 2 n + 1 } = 1 2 \begin{aligned} S & = \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) \\ & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k\pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left(\frac {1-e^{\frac {2n\pi i}{2n+1}}}{1-e^{\frac {2\pi i}{2n+1}}}\right) \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}}-e^{\pi i}}{1-e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}} + 1}{\left(1+e^{\frac {\pi i}{2n+1}} \right) \left(1-e^{\frac {\pi i}{2n+1}} \right)} \right \} \\ & = \Re \left \{ \frac 1{1-e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac 1{1-\cos \frac {\pi i}{2n+1} - i\sin \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{\left(1-\cos \frac {\pi i}{2n+1}\right)^2 + \sin^2 \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{2-2\cos \frac {\pi i}{2n+1}} \right \} \\ & = \frac 12 \end{aligned}

Chew-Seong Cheong - 3 years, 10 months ago

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Wow, so it is proved by using complex number. Thanks for explanation!

Kelvin Hong - 3 years, 10 months ago
Ahmad Saad
Jul 25, 2017

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Noel Lo
Jul 25, 2017

tan ⁡ 11 π 30 − tan ⁡ π 30 sin ⁡ 11 π 15 − sin ⁡ π 15 = sin ⁡ 11 π 30 cos ⁡ 11 π 30 − sin ⁡ π 30 cos ⁡ π 30 2 sin ⁡ 11 π 30 cos ⁡ 11 π 30 − 2 sin ⁡ π 30 cos ⁡ π 30 = sin ⁡ 11 π 30 cos ⁡ π 30 − cos ⁡ 11 π 30 sin ⁡ π 30 cos ⁡ 11 π 30 cos ⁡ π 30 ( 2 sin ⁡ 11 π 30 cos ⁡ 11 π 30 − 2 sin ⁡ π 30 cos ⁡ π 30 ) = sin ⁡ ( 11 − 1 ) π 30 2 sin ⁡ 11 π 30 cos ⁡ π 30 cos ⁡ 2 11 π 30 − 2 sin ⁡ π 30 cos ⁡ 11 π 30 cos ⁡ 2 π 30 \frac{\tan{\frac{11\pi}{30}}-\tan{\frac{\pi}{30}}}{\sin{\frac{11\pi}{15}}-\sin{\frac{\pi}{15}}}=\frac{\frac{\sin{\frac{11\pi}{30}}}{\cos{\frac{11\pi}{30}}}-\frac{\sin{\frac{\pi}{30}}}{\cos{\frac{\pi}{30}}}}{2\sin{\frac{11\pi}{30}}\cos{\frac{11\pi}{30}}-2\sin{\frac{\pi}{30}}\cos{\frac{\pi}{30}}}=\frac{\sin{\frac{11\pi}{30}}\cos{\frac{\pi}{30}}-\cos{\frac{11\pi}{30}}\sin{\frac{\pi}{30}}}{\cos{\frac{11\pi}{30}}\cos{\frac{\pi}{30}}(2\sin{\frac{11\pi}{30}}\cos{\frac{11\pi}{30}}-2\sin{\frac{\pi}{30}}\cos{\frac{\pi}{30}})}=\frac{\sin{\frac{(11-1)\pi}{30}}}{2\sin{\frac{11\pi}{30}}\cos{\frac{\pi}{30}}\cos^{2}{\frac{11\pi}{30}}-2\sin{\frac{\pi}{30}}\cos{\frac{11\pi}{30}}\cos^{2}{\frac{\pi}{30}}}

= sin ⁡ 10 π 30 ( sin ⁡ 12 π 30 + sin ⁡ 10 π 30 ) cos ⁡ 2 11 π 30 − ( sin ⁡ 12 π 30 − sin ⁡ 10 π 30 ) cos ⁡ 2 π 30 = sin ⁡ π 3 ( sin ⁡ 2 π 5 + sin ⁡ π 3 ) cos ⁡ 2 11 π 30 − ( sin ⁡ 2 π 5 − sin ⁡ π 3 ) cos ⁡ 2 π 30 = 3 2 ( sin ⁡ 2 π 5 + 3 2 ) cos ⁡ 2 11 π 30 − ( sin ⁡ 2 π 5 − 3 2 ) cos ⁡ 2 π 30 =\frac{\sin{\frac{10\pi}{30}}}{(\sin{\frac{12\pi}{30}}+\sin{\frac{10\pi}{30}})\cos^{2}{\frac{11\pi}{30}}-(\sin{\frac{12\pi}{30}}-\sin{\frac{10\pi}{30}})\cos^{2}{\frac{\pi}{30}}}=\frac{\sin{\frac{\pi}{3}}}{(\sin{\frac{2\pi}{5}}+\sin{\frac{\pi}{3}})\cos^{2}{\frac{11\pi}{30}}-(\sin{\frac{2\pi}{5}}-\sin{\frac{\pi}{3}})\cos^{2}{\frac{\pi}{30}}}=\frac{{\frac{\sqrt{3}}{2}}}{(\sin{\frac{2\pi}{5}}+{\frac{\sqrt{3}}{2}})\cos^{2}{\frac{11\pi}{30}}-(\sin{\frac{2\pi}{5}}-{\frac{\sqrt{3}}{2}})\cos^{2}{\frac{\pi}{30}}}

= 3 2 sin ⁡ 2 π 5 ( cos ⁡ 2 11 π 30 − cos ⁡ 2 π 30 ) + 3 2 ( cos ⁡ 2 11 π 30 + cos ⁡ 2 π 30 ) = 3 2 sin ⁡ 2 π 5 ( cos ⁡ 2 11 π 30 − cos ⁡ 2 π 30 ) + 3 ( cos ⁡ 2 11 π 30 + cos ⁡ 2 π 30 ) =\frac{{\frac{\sqrt{3}}{2}}}{\sin{\frac{2\pi}{5}}(\cos^{2}{\frac{11\pi}{30}}-\cos^{2}{\frac{\pi}{30}})+{\frac{\sqrt{3}}{2}}(\cos^{2}{\frac{11\pi}{30}}+\cos^{2}{\frac{\pi}{30}})}=\frac{\sqrt{3}}{2\sin{\frac{2\pi}{5}}(\cos^{2}{\frac{11\pi}{30}}-\cos^{2}{\frac{\pi}{30}})+\sqrt{3}(\cos^{2}{\frac{11\pi}{30}}+\cos^{2}{\frac{\pi}{30}})}

= 3 2 sin ⁡ 2 π 5 ( cos ⁡ 11 π 30 − cos ⁡ π 30 ) ( cos ⁡ 11 π 30 + cos ⁡ π 30 ) + 3 ( ( cos ⁡ 11 π 30 + cos ⁡ π 30 ) 2 − 2 cos ⁡ 11 π 30 cos ⁡ π 30 ) =\frac{\sqrt{3}}{2\sin{\frac{2\pi}{5}}(\cos{\frac{11\pi}{30}}-\cos{\frac{\pi}{30}})(\cos{\frac{11\pi}{30}}+\cos{\frac{\pi}{30}})+\sqrt{3}((\cos{\frac{11\pi}{30}}+\cos{\frac{\pi}{30}})^{2}-2\cos{\frac{11\pi}{30}}\cos{\frac{\pi}{30}})}

= 3 2 sin ⁡ 2 π 5 ( − 2 sin ⁡ ( 1 2 12 π 30 ) sin ⁡ ( 1 2 10 π 30 ) ) ( 2 cos ⁡ ( 1 2 12 π 30 ) cos ⁡ ( 1 2 10 π 30 ) ) + 3 ( 4 cos ⁡ 2 6 π 30 cos ⁡ 2 5 π 30 − cos ⁡ 12 π 30 − cos ⁡ 10 π 30 ) = 3 − 2 sin ⁡ 2 2 π 5 sin ⁡ π 3 + 3 ( 4 cos ⁡ 2 π 5 cos ⁡ 2 π 6 − cos ⁡ 2 π 5 − cos ⁡ π 3 ) =\frac{\sqrt{3}}{2\sin{\frac{2\pi}{5}}(-2\sin{(\frac{1}{2}\frac{12\pi}{30})}\sin{(\frac{1}{2}\frac{10\pi}{30}))}(2\cos{(\frac{1}{2}\frac{12\pi}{30})}\cos{(\frac{1}{2}\frac{10\pi}{30})})+\sqrt{3}(4\cos^{2}{\frac{6\pi}{30}}\cos^{2}{\frac{5\pi}{30}}-\cos{\frac{12\pi}{30}}-\cos{\frac{10\pi}{30}})}=\frac{\sqrt{3}}{-2\sin^{2}{\frac{2\pi}{5}}\sin{\frac{\pi}{3}}+\sqrt{3}(4\cos^{2}{\frac{\pi}{5}}\cos^{2}{\frac{\pi}{6}}-\cos{\frac{2\pi}{5}}-\cos{\frac{\pi}{3}})}

= 3 − 3 sin ⁡ 2 2 π 5 + 3 ( 3 cos ⁡ 2 π 5 − cos ⁡ 2 π 5 − 1 2 ) = 1 − sin ⁡ 2 2 π 5 + 3 cos ⁡ 2 π 5 − cos ⁡ 2 π 5 − 1 2 = 1 cos ⁡ 4 π 5 − 1 2 + 3 cos ⁡ 2 π 5 − 2 cos ⁡ 2 π 5 + 1 − 1 2 = 1 1 2 cos ⁡ 4 π 5 + cos ⁡ 2 π 5 = 1 − 1 2 cos ⁡ π 5 + cos ⁡ 2 π 5 =\frac{\sqrt{3}}{-\sqrt{3}\sin^{2}{\frac{2\pi}{5}}+\sqrt{3}(3\cos^{2}{\frac{\pi}{5}}-\cos{\frac{2\pi}{5}}-\frac{1}{2})}=\frac{1}{-\sin^{2}{\frac{2\pi}{5}}+3\cos^{2}{\frac{\pi}{5}}-\cos{\frac{2\pi}{5}}-\frac{1}{2}}=\frac{1}{\frac{\cos{\frac{4\pi}{5}}-1}{2}+3\cos^{2}{\frac{\pi}{5}}-2\cos^{2}{\frac{\pi}{5}}+1-\frac{1}{2}}=\frac{1}{\frac{1}{2}\cos{\frac{4\pi}{5}}+\cos^{2}{\frac{\pi}{5}}}=\frac{1}{-\frac{1}{2}\cos{\frac{\pi}{5}}+\cos^{2}{\frac{\pi}{5}}}

= 4 sin ⁡ π 5 4 sin ⁡ π 5 cos ⁡ 2 π 5 − 2 sin ⁡ π 5 cos ⁡ π 5 = 4 sin ⁡ π 5 2 sin ⁡ 2 π 5 cos ⁡ π 5 − sin ⁡ 2 π 5 = 4 sin ⁡ π 5 sin ⁡ ( 2 + 1 ) π 5 + sin ⁡ ( 2 − 1 ) π 5 − sin ⁡ 2 π 5 = 4 sin ⁡ π 5 sin ⁡ 3 π 5 + sin ⁡ π 5 − sin ⁡ 2 π 5 = 4 sin ⁡ π 5 sin ⁡ ( 5 − 3 ) π 5 + sin ⁡ π 5 − sin ⁡ 2 π 5 =\frac{4\sin{\frac{\pi}{5}}}{4\sin{\frac{\pi}{5}}\cos^{2}{\frac{\pi}{5}}-2\sin{\frac{\pi}{5}}\cos{\frac{\pi}{5}}}=\frac{4\sin{\frac{\pi}{5}}}{2\sin{\frac{2\pi}{5}}\cos{\frac{\pi}{5}}-\sin{\frac{2\pi}{5}}}=\frac{4\sin{\frac{\pi}{5}}}{\sin{\frac{(2+1)\pi}{5}}+\sin{\frac{(2-1)\pi}{5}}-\sin{\frac{2\pi}{5}}}=\frac{4\sin{\frac{\pi}{5}}}{\sin{\frac{3\pi}{5}}+\sin{\frac{\pi}{5}}-\sin{\frac{2\pi}{5}}}=\frac{4\sin{\frac{\pi}{5}}}{\sin{\frac{(5-3)\pi}{5}}+\sin{\frac{\pi}{5}}-\sin{\frac{2\pi}{5}}}

= 4 sin ⁡ π 5 sin ⁡ 2 π 5 + sin ⁡ π 5 − sin ⁡ 2 π 5 = 4 sin ⁡ π 5 sin ⁡ π 5 = 4 =\frac{4\sin{\frac{\pi}{5}}}{\sin{\frac{2\pi}{5}}+\sin{\frac{\pi}{5}}-\sin{\frac{2\pi}{5}}}=\frac{4\sin{\frac{\pi}{5}}}{\sin{\frac{\pi}{5}}}=\boxed{4}

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S = tan ⁡ 6 6 ∘ − tan ⁡ 6 ∘ sin ⁡ 13 2 ∘ − sin ⁡ 1 2 ∘ = sin ⁡ 6 6 ∘ cos ⁡ 6 6 ∘ − sin ⁡ 6 ∘ cos ⁡ 6 ∘ 2 cos ⁡ 7 2 ∘ sin ⁡ 6 0 ∘ = sin ⁡ 6 6 ∘ cos ⁡ 6 ∘ − cos ⁡ 6 6 ∘ sin ⁡ 6 ∘ 2 cos ⁡ 6 6 ∘ cos ⁡ 6 ∘ cos ⁡ 7 2 ∘ sin ⁡ 6 0 ∘ = sin ⁡ 6 0 ∘ ( cos ⁡ 7 2 ∘ + cos ⁡ 6 0 ∘ ) cos ⁡ 7 2 ∘ sin ⁡ 6 0 ∘ = 1 cos ⁡ 2 7 2 ∘ + 1 2 cos ⁡ 7 2 ∘ = 4 4 cos ⁡ 2 7 2 ∘ + 2 cos ⁡ 7 2 ∘ = 4 1 = 4 \begin{aligned} S &= \dfrac{\tan 66^\circ - \tan 6^\circ}{\sin 132^\circ - \sin 12^\circ}\\ &= \dfrac{\frac{\sin 66^\circ}{\cos 66^\circ} - \frac{\sin 6^\circ}{\cos 6^\circ}}{2\cos 72^\circ \sin 60^\circ}\\ &= \dfrac{\sin 66^\circ \cos 6^\circ - \cos 66^\circ \sin 6^\circ}{2 \cos 66^\circ \cos 6^\circ \cos 72^\circ \sin 60^\circ}\\ &= \dfrac{\sin 60^\circ}{(\cos 72^\circ+\cos 60^\circ) \cos 72^\circ \sin 60^\circ}\\ &= \dfrac{1}{\cos^2 72^\circ + \frac{1}{2}\cos 72^\circ}\\ &= \dfrac{4}{4 \cos^2 72^\circ+2\cos 72^\circ}\\ &= \dfrac{4}{1}\\ &= \boxed{4} \end{aligned}

Note that since cos ⁡ 7 2 ∘ = 5 − 1 4 \cos 72^\circ=\dfrac{\sqrt{5}-1}{4} , then 4 cos ⁡ 7 2 ∘ + 1 = 5 4\cos 72^\circ + 1=\sqrt{5} , so 16 cos ⁡ 2 7 2 ∘ + 8 cos ⁡ 7 2 ∘ + 1 = 5 16\cos^2 72^\circ+8\cos 72^\circ+1=5 , and 4 cos ⁡ 2 7 2 ∘ + 2 cos ⁡ 7 2 ∘ = 1 4\cos^2 72^\circ+2\cos 72^\circ=1 .

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